# Network analysis by van valkenburg solutions CH#6 part 1

September 11, 2017 | Author: Husnain Ahmad | Category: Initial Condition, Equations, Differential Calculus, Electrical Engineering, Electromagnetism

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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CH#6 PROBLEMS

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Q#6.1: Show that i = ke-2t and i = ke-t are solutions of the differential equation d2i di +3 + 2i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 a 1 b 3 c 2

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -3 ± √32 – 4(1)(2) s1, s2 = 2(1) -3 ± √9 – 8 s1, s2 = 2(1) -3 ± √1 s1, s2 = 2(1) -3 ± 1 s1, s2 = 2 s1, s2 = (-3 + 1)/2, (-3 - 1)/2 s1, s2 = (-2)/2, (-4)/2 s1, s2 = -1, -2 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-1t + K2e-2t Q#6.2: Show that i = Ke-t and i = Kte-t are solutions of the differential equation d2i di +2 +i=0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here 1s2i + 2si + 1i = 0

2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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s2 + 2s + 1 = 0 a b c

s1, s2 =

1 2 1

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-2 ± √22 – 4(1)(1) 2(1) -2 ± √4 – 4 2(1) -2 ± √0 2(1) -2 ± 0

2 s1, s2 = (-2 + 0)/2, (-2 - 0)/2 s1, s2 = (-1), (-1) General solution: i(t) = K1es1t + K2tes2t Putting corresponding values we get i(t) = K1e-1t + K2te-1t Q#6.3: Find the general solution of each of the following equations: (a) d2i di +3 + 2i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 a 1 b 3 c 2

s1, s2 =

-b ± √b2 – 4ac

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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2a Putting corresponding values we get -3 ± √32 – 4(1)(2) s1, s2 = 2(1) -3 ± √9 – 8 s1, s2 = 2(1) -3 ± √1 s1, s2 = 2(1) -3 ± 1 s1, s2 = 2 s1, s2 = (-3 + 1)/2, (-3 - 1)/2 s1, s2 = (-2)/2, (-4)/2 s1, s2 = -1, -2 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-1t + K2e-2t (b) d2i

di +5 + 6i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 5si + 6i = 0 s2 + 5s + 6 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 =

-5 ± √52 – 4(1)(6) 2(1) -5 ± √25 – 24

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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s1, s2 = s1, s2 = s1, s2 =

2(1) -5 ± √1 2(1) -5 ± 1

2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-2t + K2e-3t (c) d2i

di +7 + 12i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 7si + 12i = 0 s2 + 7s + 12 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -7 ± √72 – 4(1)(12) s1, s2 = 2(1) -7 ± √49 – 48 s1, s2 = 2(1) s1, s2 = s1, s2 =

-7 ± √1 2(1) -7 ± 1

1 7 12

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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2 s1, s2 = (-7 + 1)/2, (-7 - 1)/2 s1, s2 = (-6)/2, (-8)/2 s1, s2 = -3, -4 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-3t + K2e-4t (d) d2i

di +5 + 4i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 5si + 4i = 0 s2 + 5s + 4 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-5 ± √52 – 4(1)(4) 2(1) -5 ± √25 – 16 2(1) -5 ± √9 2(1) -5 ± 3

2 s1, s2 = (-5 + 3)/2, (-5 - 3)/2 s1, s2 = (-2)/2, (-8)/2 s1, s2 = -1, -4 General solution: i(t) = K1es1t + K2es2t

1 5 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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Putting corresponding values we get i(t) = K1e-1t + K2e-4t (e) d2i

di

+1 + 6i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 1si + 6i = 0 s2 + 1s + 6 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -1 ± √12 – 4(1)(6) s1, s2 = 2(1) -1 ± √1 – 24 s1, s2 = 2(1) -1 ± √-23 s1, s2 = 2(1) -1 ± i4.796 s1, s2 = 2 s1, s2 = (-1 + i4.796)/2, (-1 - i4.796)/2 s1, s2 = (-0.5 + i2.398), (-0.5 - i2.398) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t (f) d2i di +1 + 2i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0

1 1 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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Here s2i + 1si + 2i = 0 s2 + 1s + 2 = 0 A B C

s1, s2 =

1 1 2

-b ± √b2 – 4ac

2a Putting corresponding values we get -1 ± √12 – 4(1)(2) s1, s2 = 2(1) -1 ± √1 – 8 s1, s2 = 2(1) -1 ± √-7 s1, s2 = 2(1) -1 ± i2.646 s1, s2 = 2 s1, s2 = (-1 + i2.646)/2, (-1 - i2.646)/2 s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t (g) d2i

di +2 +i=0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 2si + i = 0 s2 + 2s + 1 = 0 a b c 2 -b ± √b – 4ac s1, s2 = 2a

1 2 1

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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Putting corresponding values we get -2 ± √22 – 4(1)(1) s1, s2 = 2(1) -2 ± √4 – 4 s1, s2 = 2(1) -2 ± √0 s1, s2 = 2(1) -2 ± 0 s1, s2 = 2 s1, s2 = (-2 + 0)/2, (-2 - 0)/2 s1, s2 = (-1), (-1) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2te(-1)t (h) d2i

di +4 + 4i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 4si + 4i = 0 s2 + 4s + 4 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -4 ± √42 – 4(1)(4) s1, s2 = 2(1) -4 ± √16 – 16 s1, s2 = 2(1) -4 ± √0

1 4 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} s1, s2 = s1, s2 =

2(1) -4 ± 0

2 s1, s2 = (-4 + 0)/2, (-4 - 0)/2 s1, s2 = (-2), (-2) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-2)t + K2te(-2)t Q#6.4: Find the general solution of each of the following homogeneous differential equations: (a) d2v dv +2 + 2v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 2v = 0 s2 + 2s + 2 = 0 a 1 b 2 c 2

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2 ± √22 – 4(1)(2) s1, s2 = 2(1)

s1, s2 = s1, s2 = s1, s2 =

-2 ± √4 – 8 2(1) -2 ± √-4 2(1) -2 ± i2

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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2 s1, s2 = (-2 + i2)/2, (-2 – i2)/2 s1, s2 = (-1 + i1), (-1 – i1) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1)t + K2e(-1 – i1)t (b) d2v

dv +2 + 4v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 4v = 0 s2 + 2s + 4 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 = s1, s2 =

-2 ± √22 – 4(1)(4) 2(1) -2 ± √4 – 16 2(1)

s1, s2 = s1, s2 =

-2 ± √-12 2(1) -2 ± i3.464

2 s1, s2 = (-2 + i3.464)/2, (-2 – i3.464)/2 s1, s2 = (-1 + i1.732), (-1 – i1.732) General solution:

1 2 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1.732)t + K2e(-1 – i1.732)t (c) d2v

dv +4 + 2v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 4sv + 2v = 0 s2 + 4s + 2 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-4 ± √42 – 4(1)(2) 2(1) -4 ± √16 – 8 2(1) -4 ± √8 2(1) -4 ± 2.828

2 s1, s2 = (-4 + 2.828)/2, (-4 – 2.828)/2 s1, s2 = (-0.586), (-3.414) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-0.586)t + K2e(-3.414)t (d) d2v dv 2 +8 + 16v = 0

1 4 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} dt2 dt After simplification d2v dv +4 + 8v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 4sv + 8v = 0 s2 + 4s + 8 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 =

-4 ± √42 – 4(1)(8) 2(1) -4 ± √16 – 32 2(1) -4 ± √-16 2(1) -4 ± i4

s1, s2 =

2 s1, s2 = (-4 + i4)/2, (-4 – i4)/2 s1, s2 = (-2 + i2), (-2 – i2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-2 + i2)t + K2e(-2 – i2)t (e) d2v

dv

+2 + 3v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here

13

1 4 8

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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s2v + 2sv + 3v = 0 s2 + 2s + 3 = 0 a b c

s1, s2 =

1 2 3

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 =

-2 ± √22 – 4(1)(3) 2(1) -2 ± √4 – 12 2(1) -2 ± √-8 2(1) -2 ± i2.828

s1, s2 =

2 s1, s2 = (-2 + i2.828)/2, (-2 – i2.828)/2 s1, s2 = (-1 + i1.414), (-1 – i1.414) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1.414)t + K2e(-1 – i1.414)t

(f) d2v

dv

+3 + 5v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2v + 3sv + 5v = 0 s2 + 3s + 5 = 0 a b c

1 3 5

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

15

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-3 ± √32 – 4(1)(5) 2(1) -3 ± √9 – 20 2(1) -3 ± √-11 2(1) -3 ± i3.317

2 s1, s2 = (-3 + i3.317)/2, (-3 – i3.317)/2 s1, s2 = (-1.5 + i1.659), (-1.5 – i1.659) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1.5 + i1.659)t + K2e(-1.5 – i1.659)t Q#6.5: Find particular solutions for the differential equations of Prob. 6.3 subject to the initial conditions: di i(0+) = 1, (0+) = 0 dt (a) d2i di +3 + 2i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 a 1 b 3 c 2

s1, s2 =

-b ± √b2 – 4ac

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 2a Putting corresponding values we get -3 ± √32 – 4(1)(2) s1, s2 = 2(1) -3 ± √9 – 8 s1, s2 = 2(1) -3 ± √1 s1, s2 = 2(1) -3 ± 1 s1, s2 = 2 s1, s2 = (-3 + 1)/2, (-3 - 1)/2 s1, s2 = (-2)/2, (-4)/2 s1, s2 = -1, -2 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-1t + K2e-2t … (i) At t = 0+ i(0+) = K1e-1(0+) + K2e-2(0+) 1 = K1e0 + K2e0 1 = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’

di(t) = -K1e-1t - 2K2e-2t dt di(0+) = -K1e-1(0+) - 2K2e-2(0+) dt di(0+) = -K1e0 - 2K2e0 dt 0 = -K1(1) - 2K2(1) 0 = -K1 - 2K2 K1 = - 2K2 Putting the value of K1 in (ii) 1 = -2K2 + K2

16

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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K2 = -1 Putting the value of K2 in (ii) 1 = K1 + (-1) K1 = 2 Putting corresponding values we get i(t) = 2e-t + (-1)e-2t i(t) = 2e-t – e-2t (b) d2i

di +5 + 6i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 5si + 6i = 0 s2 + 5s + 6 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-5 ± √52 – 4(1)(6) 2(1) -5 ± √25 – 24 2(1) -5 ± √1 2(1) -5 ± 1

2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} s1, s2 = -2, -3 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-2t + K2e-3t … (i) At t = 0+ i(0+) = K1e-2(0+) + K2e-3(0+) i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = -2K1e-2t - 3K2e-3t dt di(0+) = -2K1e-2(0+) - 3K2e-3(0+) dt di(0+) = -2K1e0 - 3K2e0 dt di(0+) = -2K1(1) - 3K2(1) dt 0 = -2K1 - 3K2 2K1 = -3K2 K1 = -1.5K2 1 = K1 + K2 … (ii) Putting the value of K1 in (ii) 1 = -1.5K2 + K2 1 = -0.5K2 K2 = -2 K1 = -1.5(-2) K1 = 3.0 Putting corresponding values we get i(t) = 3e-2t + (-2)e-3t i(t) = 3e-2t - 2e-3t (c) d2i

di +7 + 12i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0

18

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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Here s2i + 7si + 12i = 0 s2 + 7s + 12 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -7 ± √72 – 4(1)(12) s1, s2 = 2(1) -7 ± √49 – 48 s1, s2 = 2(1) s1, s2 = s1, s2 =

-7 ± √1 2(1) -7 ± 1

2 s1, s2 = (-7 + 1)/2, (-7 - 1)/2 s1, s2 = (-6)/2, (-8)/2 s1, s2 = -3, -4 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-3t + K2e-4t … (i) At t = 0+ i(0+) = K1e-3(0+) + K2e-4(0+) i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = -3K1e-3t - 4K2e-4t dt di(0+) = -3K1e-3(0+) - 4K2e-4(0+) dt di(0+)

1 7 12

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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= -3K1e0 - 4K2e0 dt di(0+) = -3K1(1) - 4K2(1) dt di(0+) = -3K1 - 4K2 dt 0 = -3K1 - 4K2 3K1 = - 4K2 K1 = -1.334K2 Putting the value of K1 in (ii) 1 = -1.334K2 + K2 1 = -0.334K2 K2 = -2.994 K1 = -1.334(-2.994) = 3.994 Putting corresponding values we get i(t) = (3.994)e-3t + (-2.994)e-4t i(t) = 3.994e-3t - 2.994e-4t (d) d2i di +5 + 4i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 5si + 4i = 0 s2 + 5s + 4 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 =

-5 ± √52 – 4(1)(4) 2(1) -5 ± √25 – 16 2(1) -5 ± √9

1 5 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

2(1) -5 ± 3

2 s1, s2 = (-5 + 3)/2, (-5 - 3)/2 s1, s2 = (-2)/2, (-8)/2 s1, s2 = -1, -4 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-1t + K2e-4t … (i) At t = 0+ i(0+) = K1e-1(0+) + K2e-4(0+) i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = -1K1e-1t - 4K2e-4t dt di(0+) = -1K1e-1(0+) - 4K2e-4(0+) dt di(0+) = -1K1e0 - 4K2e0 dt 0 = -1K1(1) - 4K2(1) 0 = -K1 - 4K2 K1 = -4K2 Putting the value of K1 in (ii) 1 = -4K2 + K2 1 = -3K2 K2 = -0.334 K1 = -4(-0.334) K1 = 1.336 Putting corresponding values we get i(t) = 1.336e-1t – 0.334e-4t (e) d2i

di +1 + 6i = 0 2 dt dt Characteristic equation:

21

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

22

as2 + bs + c = 0 Here s2i + 1si + 6i = 0 s2 + 1s + 6 = 0 A B C

s1, s2 =

1 1 6

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 =

-1 ± √12 – 4(1)(6) 2(1) -1 ± √1 – 24 2(1) -1 ± √-23 2(1)

s1, s2 =

-1 ± i4.796

2 s1, s2 = (-1 + i4.796)/2, (-1 - i4.796)/2 s1, s2 = (-0.5 + i2.398), (-0.5 - i2.398) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t … (i) At t = 0+ i(0+) = K1e(-0.5 + i2.398)0+ + K2e(-0.5 - i2.398)0+ i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t dt di(0+) = (-0.5 + i2.398)K1e(-0.5 + i2.398)0+ + (-0.5 - i2.398)K2e(-0.5 - i2.398)0+

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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dt di(0+) = (-0.5 + i2.398)K1e0 + (-0.5 - i2.398)K2e0 dt 0 = (-0.5 + i2.398)K1(1) + (-0.5 - i2.398)K2(1) 0 = (-0.5 + i2.398)K1 + (-0.5 - i2.398)K2 0.5 + i2.398 K1 = K2 -0.5 + i2.398 K1 = [0.917 – i0.400]K2 Putting the value of K1 in (ii) 1 = [0.917 – i0.400]K2 + K2 1 = [[0.917 – i0.400] + 1]K2 1 = [0.917 – i0.400 + 1]K2 1 = [1.917 – i0.400]K2 K2 = 0.5 + i0.104 1 = K1 + K2 K1 = 1 – K2 K1 = 1 – [0.5 + i0.104] K1 = 1 – 0.5 - i0.104 K1 = 0.5 - i0.104 Putting corresponding values we get i(t) = [0.5 - i0.104]e(-0.5 + i2.398)t + [0.5 + i0.104]e(-0.5 - i2.398)t (f) d2i di +1 + 2i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 1si + 2i = 0 s2 + 1s + 2 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -1 ± √12 – 4(1)(2) s1, s2 = 2(1) -1 ± √1 – 8 s1, s2 =

1 1 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 = s1, s2 =

2(1) -1 ± √-7 2(1) -1 ± i2.646

2 s1, s2 = (-1 + i2.646)/2, (-1 - i2.646)/2 s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t … (i) At t = 0+ i(0+) = K1e(-0.5 + i1.323)0+ + K2e(-0.5 - i1.323)0+ i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t dt di(0+) = (-0.5 + i1.323)K1e(-0.5 + i1.323)0+ + (-0.5 - i1.323)K2e(-0.5 - i1.323)0+ dt di(0+) = (-0.5 + i1.323)K1e0 + (-0.5 - i1.323)K2e0 dt di(0+) = (-0.5 + i1.323)K1(1) + (-0.5 - i1.323)K2(1) dt 0 = (-0.5 + i1.323)K1 + (-0.5 - i1.323)K2 (-0.5 + i1.323)K1 = (0.5 + i1.323)K2 (0.5 + i1.323) K1 = K2 (-0.5 + i1.323) K1 = (0.75 – 0.662i)K2 1 = (0.75 – 0.662i)K2 + K2 1 = [(0.75 – 0.662i) + 1]K2 1 = [1.75 – 0.662i]K2

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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K2 = 0.5 + 0.189i Putting corresponding values we get 1 = K1 + 0.5 + 0.189i K1 = 1 - 0.5 - 0.189i K1 = 0.5 - 0.189i Putting corresponding values we get i(t) = [0.5 - 0.189i]e(-0.5 + i1.323)t + [0.5 + 0.189i]e(-0.5 - i1.323)t (g) d2i

di +2 +i=0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 2si + i = 0 s2 + 2s + 1 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2 ± √22 – 4(1)(1) s1, s2 = 2(1) s1, s2 = s1, s2 = s1, s2 =

-2 ± √4 – 4 2(1) -2 ± √0 2(1) -2 ± 0

2 s1, s2 = (-2 + 0)/2, (-2 - 0)/2 s1, s2 = (-1), (-1) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get

1 2 1

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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i(t) = K1e(-1)t + K2te(-1)t … (i) At t = 0+ i(t) = K1e(-1)t + K2te(-1)t i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+ i(0+) = K1e0 i(0+) = K1(1) 1 = K1 Differentiating (i) with respect to ‘t’ di(t) = -K1e(-1)t + K2[-te(-1)t – e-t] dt di(0+) = (-1)K1e(-1)0+ + K2[-(0+)e-(0+) – e-(0+)] dt di(0+) = (-1)K1e0 + K2[–e0] dt di(0+) = (-1)K1(1) + K2[–(1)] dt 0 = -K1 – K2 0 = -(1) – K2 K2 = -1 Putting corresponding values i(t) = (1)e(-1)t + (-1)te(-1)t i(t) = e(-1)t - te(-1)t (h) d2i

di

+4 + 4i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 4si + 4i = 0 s2 + 4s + 4 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

1 4 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 2a Putting corresponding values we get -4 ± √42 – 4(1)(4) s1, s2 = 2(1) -4 ± √16 – 16 s1, s2 = 2(1) -4 ± √0 s1, s2 = 2(1) -4 ± 0 s1, s2 = 2 s1, s2 = (-4 + 0)/2, (-4 - 0)/2 s1, s2 = (-2), (-2) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-2)t + K2te(-2)t At t = 0+ i(0+) = K1e(-2)0+ + K2(0+)e(-2)0+ 1 = K1e0 1 = K1(1) 1 = K1 Differentiating (i) with respect to ‘t’ di(t) = -2K1e(-2)t + K2[-0.5te-2t – 0.25e-2t] dt di(0+) = -2K1e(-2)0+ + K2[-0.5(0+)e-2(0+) – 0.25e-2(0+)] dt di(0+) = -2K1e0 + K2[–0.25e0] dt di(0+) = -2K1(1) + K2[–0.25(1)] dt 0 = -2K1 – 0.25K2 2K1 = -0.25K2 K1 = -0.125K2 K2 = -8 Putting corresponding values i(t) = (1)e(-2)t + (-8)te(-2)t

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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i(t) = e(-2)t - 8te(-2)t Q#6.6: Find particular solutions for the differential equations of Prob. 6.3 subject to the initial conditions: di i(0+) = 2, (0+) = +1 dt (a) d2i di +3 + 2i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 a 1 b 3 c 2

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -3 ± √32 – 4(1)(2) s1, s2 = 2(1) -3 ± √9 – 8 s1, s2 = 2(1) -3 ± √1 s1, s2 = 2(1) -3 ± 1 s1, s2 = 2 s1, s2 = (-3 + 1)/2, (-3 - 1)/2 s1, s2 = (-2)/2, (-4)/2 s1, s2 = -1, -2 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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i(t) = K1e-1t + K2e-2t … (i) At t = 0+ i(0+) = K1e-1(0+) + K2e-2(0+) 2 = K1e0 + K2e0 2 = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = -K1e-1t - 2K2e-2t dt di(0+) = -K1e-1(0+) - 2K2e-2(0+) dt di(0+) = -K1e0 - 2K2e0 dt 1 = -K1(1) - 2K2(1) 1 = -K1 - 2K2 K1 = -1 - 2K2 Putting the value of K1 in (ii) 2 = [-1 - 2K2] + K2 2 = -1 - 2K2 + K2 3 = -K2 K2 = -3 Putting the value of K2 in (ii) 2 = K1 + (-3) K1 = 5 Putting corresponding values we get i(t) = 5e-t + (-3)e-2t i(t) = 5e-t – 3e-2t (b) d2i

di

+5 + 6i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 5si + 6i = 0 s2 + 5s + 6 = 0 A B C

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -5 ± √52 – 4(1)(6) s1, s2 = 2(1) -5 ± √25 – 24 s1, s2 = 2(1) -5 ± √1 s1, s2 = 2(1) -5 ± 1 s1, s2 = 2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-2t + K2e-3t … (i) At t = 0+ i(0+) = K1e-2(0+) + K2e-3(0+) i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = -2K1e-2t - 3K2e-3t dt di(0+) = -2K1e-2(0+) - 3K2e-3(0+) dt di(0+) = -2K1e0 - 3K2e0 dt di(0+) = -2K1(1) - 3K2(1) dt 1 = -2K1 - 3K2 2K1 = -3K2 - 1 K1 = -1.5K2 – 0.5

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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2 = K1 + K2 … (ii) Putting the value of K1 in (ii) 2 = [-1.5K2 – 0.5] + K2 2 = -1.5K2 – 0.5 + K2 2 = -0.5K2 – 0.5 K2 = -5 2 = K1 + (-5) K1 = 7 Putting corresponding values we get i(t) = 7e-2t + (-5)e-3t i(t) = 7e-2t - 5e-3t (c) d2i

di

+7 + 12i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 7si + 12i = 0 s2 + 7s + 12 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -7 ± √72 – 4(1)(12) s1, s2 = 2(1) -7 ± √49 – 48 s1, s2 = 2(1) s1, s2 = s1, s2 =

-7 ± √1 2(1) -7 ± 1

2 s1, s2 = (-7 + 1)/2, (-7 - 1)/2 s1, s2 = (-6)/2, (-8)/2 s1, s2 = -3, -4

1 7 12

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-3t + K2e-4t … (i) At t = 0+ i(0+) = K1e-3(0+) + K2e-4(0+) i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = -3K1e-3t - 4K2e-4t dt di(0+) = -3K1e-3(0+) - 4K2e-4(0+) dt

di(0+) = -3K1e0 - 4K2e0 dt di(0+) = -3K1(1) - 4K2(1) dt di(0+) = -3K1 - 4K2 dt 1 = -3K1 - 4K2 3K1 = -4K2 - 1 K1 = -1.334K2 – 0.334 Putting the value of K1 in (ii) 2 = [-1.334K2 – 0.334] + K2 2 = -1.334K2 – 0.334 + K2 2.334 = -0.334K2 K2 = -6.988 2 = K1 + (-6.988) 8.988 = K1 Putting corresponding values we get i(t) = (8.988)e-3t + (-6.988)e-4t i(t) = 8.988e-3t – 6.988e-4t (d)

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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d2i

di +5 + 4i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 5si + 4i = 0 s2 + 5s + 4 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 = s1, s2 =

-5 ± √52 – 4(1)(4) 2(1) -5 ± √25 – 16 2(1)

s1, s2 = s1, s2 =

-5 ± √9 2(1) -5 ± 3

2 s1, s2 = (-5 + 3)/2, (-5 - 3)/2 s1, s2 = (-2)/2, (-8)/2 s1, s2 = -1, -4 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-1t + K2e-4t … (i) At t = 0+ i(0+) = K1e-1(0+) + K2e-4(0+) i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 2 = K1 + K2 … (ii)

1 5 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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Differentiating (i) with respect to ‘t’ di(t) = -1K1e-1t - 4K2e-4t dt di(0+) = -1K1e-1(0+) - 4K2e-4(0+) dt di(0+) = -1K1e0 - 4K2e0 dt 0 = -1K1(1) - 4K2(1) 0 = -K1 - 4K2 K1 = -4K2 Putting the value of K1 in (ii) 1 = -4K2 + K2 1 = -3K2 K2 = -0.334 K1 = -4(-0.334) K1 = 1.336 Putting corresponding values we get i(t) = 1.336e-1t – 0.334e-4t (e) d2i

di +1 + 6i = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 1si + 6i = 0 s2 + 1s + 6 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 =

-1 ± √12 – 4(1)(6) 2(1) -1 ± √1 – 24

1 1 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

2(1) -1 ± √-23 2(1) -1 ± i4.796

s1, s2 =

2 s1, s2 = (-1 + i4.796)/2, (-1 - i4.796)/2 s1, s2 = (-0.5 + i2.398), (-0.5 - i2.398) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t … (i) At t = 0+ i(0+) = K1e(-0.5 + i2.398)0+ + K2e(-0.5 - i2.398)0+ i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t dt di(0+) = (-0.5 + i2.398)K1e(-0.5 + i2.398)0+ + (-0.5 - i2.398)K2e(-0.5 - i2.398)0+ dt di(0+) = (-0.5 + i2.398)K1e0 + (-0.5 - i2.398)K2e0 dt 1 = (-0.5 + i2.398)K1(1) + (-0.5 - i2.398)K2(1) 1 = (-0.5 + i2.398)K1 + (-0.5 - i2.398)K2 After simplification K1 = -0.084 – i0.400 + [0.917 – i0.4]K2 2 = [-0.084 – i0.400 + [0.917 – i0.4]K2] + K2 2 + 0.084 + i0.400 = [1.917 – i0.4]K2 2.084 + i0.400 = [1.917 – i0.4]K2 K2 = 1 + i0.417 2 = K1 + 1 + i0.417 2 - 1 - i0.417 = K1 1 - i0.417 = K1 Putting corresponding values we get i(t) = [1 - i0.417]e(-0.5 + i2.398)t + [1 + i0.417]e(-0.5 - i2.398)t (f) d2i

di

35

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} +1 + 2i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 1si + 2i = 0 s2 + 1s + 2 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-1 ± √12 – 4(1)(2) 2(1) -1 ± √1 – 8 2(1) -1 ± √-7 2(1) -1 ± i2.646

2 s1, s2 = (-1 + i2.646)/2, (-1 - i2.646)/2 s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t … (i) At t = 0+ i(0+) = K1e(-0.5 + i1.323)0+ + K2e(-0.5 - i1.323)0+ i(0+) = K1e0 + K2e0 i(0+) = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ di(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t dt

36

1 1 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

37

di(0+) = (-0.5 + i1.323)K1e(-0.5 + i1.323)0+ + (-0.5 - i1.323)K2e(-0.5 - i1.323)0+ dt di(0+) = (-0.5 + i1.323)K1e0 + (-0.5 - i1.323)K2e0 dt di(0+) = (-0.5 + i1.323)K1(1) + (-0.5 - i1.323)K2(1) dt 1 = (-0.5 + i1.323)K1 + (-0.5 - i1.323)K2 K1 = -0.25 – i0.662 + [0.75 – i0.662]K2 2 = -0.25 – i0.662 + [0.75 – i0.662]K2 + K2 2 + 0.25 + i0.662 = [0.75 – i0.662]K2 + K2 2.25 + i0.662 = [1.75 – i0.662]K2 K2 = 1 + i0.757 2 = K1 + 1 + i0.757 K1 = 2 – 1 - i0.757 K1 = 1 - i0.757 i(t) = [1 - i0.757]e(-0.5 + i1.323)t + [1 + i0.757]e(-0.5 - i1.323)t (g) d2i

di +2 +i=0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2i + 2si + i = 0 s2 + 2s + 1 = 0 A B C 2 -b ± √b – 4ac s1, s2 = 2a Putting corresponding values we get -2 ± √22 – 4(1)(1) s1, s2 = 2(1) s1, s2 = s1, s2 =

-2 ± √4 – 4 2(1) -2 ± √0

1 2 1

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 2(1) -2 ± 0

s1, s2 =

2 s1, s2 = (-2 + 0)/2, (-2 - 0)/2 s1, s2 = (-1), (-1) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2te(-1)t … (i) At t = 0+ i(t) = K1e(-1)t + K2te(-1)t i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+ i(0+) = K1e0 i(0+) = K1(1) 2 = K1 Differentiating (i) with respect to ‘t’ di(t) = -K1e(-1)t + K2[-te(-1)t – e-t] dt di(0+) = (-1)K1e(-1)0+ + K2[-(0+)e-(0+) – e-(0+)] dt di(0+) = (-1)K1e0 + K2[–e0] dt di(0+) = (-1)K1(1) + K2[–(1)] dt 1 = -K1 – K2 1 = -(2) – K2 K2 = -3 Putting corresponding values i(t) = (2)e(-1)t + (-3)te(-1)t i(t) = 2e(-1)t - 3te(-1)t (h) d2i

di

+4 + 4i = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2i + 4si + 4i = 0

38

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

39

s2 + 4s + 4 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -4 ± √42 – 4(1)(4) s1, s2 = 2(1)

s1, s2 = s1, s2 = s1, s2 =

-4 ± √16 – 16 2(1) -4 ± √0 2(1) -4 ± 0

2 s1, s2 = (-4 + 0)/2, (-4 - 0)/2 s1, s2 = (-2), (-2) General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-2)t + K2te(-2)t At t = 0+ i(0+) = K1e(-2)0+ + K2(0+)e(-2)0+ 2 = K1e0 2 = K1(1) 2 = K1 Differentiating (i) with respect to ‘t’ di(t) = -2K1e(-2)t + K2[-0.5te-2t – 0.25e-2t] dt di(0+) = -2K1e(-2)0+ + K2[-0.5(0+)e-2(0+) – 0.25e-2(0+)] dt di(0+) = -2K1e0 + K2[–0.25e0]

1 4 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

40

dt di(0+) = -2K1(1) + K2[–0.25(1)] dt 1 = -2K1 – 0.25K2 1 = -2(2) – 0.25K2 -5 = 0.25K2 K2 = -20 Putting corresponding values i(t) = (2)e(-2)t + (-20)te(-2)t i(t) = 2e(-2)t - 20te(-2)t Q#6.7: Find particular solutions for the differential equations of Prob. 6.3 subject to the initial conditions: dv v(0+) = 1, (0+) = -1 dt (a) d2v dv +2 + 2v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 2v = 0 s2 + 2s + 2 = 0 a 1 b 2 c 2

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2 ± √22 – 4(1)(2) s1, s2 = 2(1) -2 ± √4 – 8 s1, s2 = 2(1) -2 ± √-4 s1, s2 = 2(1) -2 ± i2 s1, s2 =

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 2 s1, s2 = (-2 + i2)/2, (-2 – i2)/2 s1, s2 = (-1 + i1), (-1 – i1) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1)t + K2e(-1 – i1)t … (i) At t = 0+ v(0+) = K1e(-1 + i1)0+ + K2e(-1 – i1)0+ v(0+) = K1e0 + K2e0 1 = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-1 + i1)K1e(-1 + i1)t + (-1 – i1)K2e(-1 – i1)t dt dv(0+) = (-1 + i1)K1e(-1 + i1)0+ + (-1 – i1)K2e(-1 – i1)0+ dt dv(0+) = (-1 + i1)K1e0 + (-1 – i1)K2e0 dt dv(0+) = (-1 + i1)K1(1) + (-1 – i1)K2(1) dt -1 = (-1 + i1)K1 + (-1 – i1)K2 K1 = -iK2 + 0.5 + 0.5i 1 = [-iK2 + 0.5 + 0.5i] + K2 1 = -iK2 + 0.5 + 0.5i + K2 K2 = 0.5 1 = K1 + 0.5 K1 = 0.5 v(t) = 0.5e(-1 + i1)t + 0.5e(-1 – i1)t (b) d2v

dv +2 + 4v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 4v = 0 s2 + 2s + 4 = 0

41

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2 ± √22 – 4(1)(4) s1, s2 = 2(1) -2 ± √4 – 16 s1, s2 = 2(1) s1, s2 = s1, s2 =

-2 ± √-12 2(1) -2 ± i3.464

2 s1, s2 = (-2 + i3.464)/2, (-2 – i3.464)/2 s1, s2 = (-1 + i1.732), (-1 – i1.732) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1.732)t + K2e(-1 – i1.732)t … (i) At t = 0+ v(0+) = K1e(-1 + i1.732)0+ + K2e(-1 – i1.732)0+ 1 = K1e0 + K2e0 1 = K1(1) + K2(1) 1 = K1 + K2 Differentiating (i) with respect to ‘t’ dv(t) = (-1 + i1.732)K1e(-1 + i1.732)t + (-1 – i1.732)K2e(-1 – i1.732)t dt dv(0+) = (-1 + i1.732)K1e(-1 + i1.732)0+ + (-1 – i1.732)K2e(-1 – i1.732)0+ dt -1 = (-1 + i1.732)K1e0 + (-1 – i1.732)K2e0 -1 = (-1 + i1.732)K1(1) + (-1 – i1.732)K2(1) -1 = (-1 + i1.732)K1 + (-1 – i1.732)K2 (-1 + i1.732)K1 = (1 + i1.732)K2 - 1 K1 = [0.5 – i0.866]K2 + 0.25 + i0.433

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

43

1 = [[0.5 – i0.866]K2 + 0.25 + i0.433] + K2 1 = [0.5 – i0.866]K2 + 0.25 + i0.433 + K2 1 - 0.25 - i0.433 = [0.5 – i0.866]K2 + K2 0.75 - i0.433 = [0.5 – i0.866 + 1]K2 0.75 - i0.433 = [1.5 – i0.866]K2 K2 = 0.5 1 = K1 + 0.5 K1 = 0.5 Putting corresponding values v(t) = 0.5e(-1 + i1.732)t + 0.5e(-1 – i1.732)t (c) d2v

dv +4 + 2v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 4sv + 2v = 0 s2 + 4s + 2 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-4 ± √42 – 4(1)(2) 2(1) -4 ± √16 – 8 2(1) -4 ± √8 2(1) -4 ± 2.828

2 s1, s2 = (-4 + 2.828)/2, (-4 – 2.828)/2 s1, s2 = (-0.586), (-3.414) General solution: v(t) = K1es1t + K2es2t

1 4 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

44

Putting corresponding values we get v(t) = K1e(-0.586)t + K2e(-3.414)t … (i) At t = 0+ v(0+) = K1e(-0.586)0+ + K2e(-3.414)0+ 1 = K1e0 + K2e0 1 = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-0.586)K1e(-0.586)t + (-3.414)K2e(-3.414)t dt dv(0+) = (-0.586)K1e(-0.586)0+ + (-3.414)K2e(-3.414)0+ dt -1 = (-0.586)K1e0 + (-3.414)K2e0 -1 = (-0.586)K1(1) + (-3.414)K2(1) -1 = -0.586K1 - 3.414K2 K1 = 1.706 – 5.826K2 1 = 1.706 – 5.826K2 + K2 0.706 = 4.826K2 K2 = 0.146 K1 = 1.706 – 5.826(0.146) K1 = 1.706 – 0.851 K1 = 0.855 Putting corresponding values we get v(t) = 0.855e(-0.586)t + 0.146e(-3.414)t (d) d2v dv 2 +8 + 16v = 0 2 dt dt After simplification d2v dv +4 + 8v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 4sv + 8v = 0 s2 + 4s + 8 = 0 a b c

1 4 8

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 =

-4 ± √42 – 4(1)(8) 2(1)

s1, s2 = s1, s2 = s1, s2 =

-4 ± √16 – 32 2(1) -4 ± √-16 2(1) -4 ± i4

2 s1, s2 = (-4 + i4)/2, (-4 – i4)/2 s1, s2 = (-2 + i2), (-2 – i2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-2 + i2)t + K2e(-2 – i2)t … (i) At t = 0+ v(0+) = K1e(-2 + i2)0+ + K2e(-2 – i2)0+ 1 = K1e0 + K2e0 1 = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-2 + i2)K1e(-2 + i2)t + (-2 – i2)K2e(-2 – i2)t dt dv(0+) = (-2 + i2)K1e(-2 + i2)0+ + (-2 – i2)K2e(-2 – i2)0+ dt -1 = (-2 + i2)K1e0 + (-2 – i2)K2e0 -1 = (-2 + i2)K1(1) + (-2 – i2)K2(1) -1 = (-2 + i2)K1 + (-2 – i2)K2 K1 = -1i + 0.25 + i0.25 K1 = -0.75i + 0.25 1 = [-0.75i + 0.25] + K2

45

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

46

1 + 0.75i – 0.25 = K2 0.75 + 0.75i = K2 v(t) = [-0.75i + 0.25]e(-2 + i2)t + [0.75 + 0.75i]e(-2 – i2)t (e) d2v

dv +2 + 3v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 3v = 0 s2 + 2s + 3 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-2 ± √22 – 4(1)(3) 2(1) -2 ± √4 – 12 2(1) -2 ± √-8 2(1) -2 ± i2.828

2 s1, s2 = (-2 + i2.828)/2, (-2 – i2.828)/2 s1, s2 = (-1 + i1.414), (-1 – i1.414) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1.414)t + K2e(-1 – i1.414)t At t = 0+ v(0+) = K1e(-1 + i1.414)0+ + K2e(-1 – i1.414)0+ 1 = K1e0 + K2e0

1 2 3

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

47

1 = K1(1) + K2(1) 1 = K1 + K2 Differentiating (i) with respect to ‘t’ dv(t) = (-1 + i1.414)K1e(-1 + i1.414)t + (-1 – i1.414)K2e(-1 – i1.414)t dt dv(0+) = (-1 + i1.414)K1e(-1 + i1.414)0+ + (-1 – i1.414)K2e(-1 – i1.414)0+ dt dv(0+) = (-1 + i1.414)K1e0 + (-1 – i1.414)K2e0 dt dv(0+) = (-1 + i1.414)K1(1) + (-1 – i1.414)K2(1) dt -1 = (-1 + i1.414)K1 + (-1 – i1.414)K2 K1 = [0.334 – i0.943]K2 + 0.334 + i0.471 1 = [0.334 – i0.943]K2 + 0.334 + i0.471 + K2 1 - 0.334 - i0.471 = [1.334 – i0.943]K2 (0.666 - i0.471) = [1.334 – i0.943]K2 K2 = 0.499 1 = K1 + 0.499 0.501 = K1 Putting corresponding values we get v(t) = 0.501e(-1 + i1.414)t + 0.499e(-1 – i1.414)t (f) d2v

dv +3 + 5v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 3sv + 5v = 0 s2 + 3s + 5 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -3 ± √32 – 4(1)(5)

1 3 5

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} s1, s2 = s1, s2 = s1, s2 = s1, s2 =

2(1) -3 ± √9 – 20 2(1) -3 ± √-11 2(1) -3 ± i3.317

2 s1, s2 = (-3 + i3.317)/2, (-3 – i3.317)/2 s1, s2 = (-1.5 + i1.659), (-1.5 – i1.659) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1.5 + i1.659)t + K2e(-1.5 – i1.659)t … (i) At t = 0+ v(0+) = K1e(-1.5 + i1.659)0+ + K2e(-1.5 – i1.659)0+ 1 = K1e0 + K2e0 1 = K1(1) + K2(1) 1 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-1.5 + i1.659)K1e(-1.5 + i1.659)t + (-1.5 – i1.659)K2e(-1.5 – i1.659)t dt dv(0+) = (-1.5 + i1.659)K1e(-1.5 + i1.659)0+ + (-1.5 – i1.659)K2e(-1.5 – i1.659)0+ dt dv(0+) = (-1.5 + i1.659)K1e0 + (-1.5 – i1.659)K2e0 dt dv(0+) = (-1.5 + i1.659)K1(1) + (-1.5 – i1.659)K2(1) dt -1 = (-1.5 + i1.659)K1 + (-1.5 – i1.659)K2 K1 = 0.401 – i0.664 1 = 0.401 – i0.664 + K2 K2 = 0.599 + i0.664 Putting corresponding values we get v(t) = (0.401 – i0.664)e(-1.5 + i1.659)t + (0.599 + i0.664)e(-1.5 – i1.659)t

48

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Q#6.8: Find particular solutions to the differential equations given in Prob. 6.4, given the initial conditions: dv v(0+) = 2, (0+) = 1 dt (a) d2v dv +2 + 2v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 2v = 0 s2 + 2s + 2 = 0 a 1 b 2 c 2

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2 ± √22 – 4(1)(2) s1, s2 = 2(1) -2 ± √4 – 8 s1, s2 = 2(1) -2 ± √-4 s1, s2 = 2(1) -2 ± i2 s1, s2 = 2 s1, s2 = (-2 + i2)/2, (-2 – i2)/2 s1, s2 = (-1 + i1), (-1 – i1) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1)t + K2e(-1 – i1)t … (i) At t = 0+ v(0+) = K1e(-1 + i1)0+ + K2e(-1 – i1)0+ v(0+) = K1e0 + K2e0 2 = K1(1) + K2(1)

49

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

50

2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-1 + i1)K1e(-1 + i1)t + (-1 – i1)K2e(-1 – i1)t dt dv(0+) = (-1 + i1)K1e(-1 + i1)0+ + (-1 – i1)K2e(-1 – i1)0+ dt dv(0+) = (-1 + i1)K1e0 + (-1 – i1)K2e0 dt dv(0+) = (-1 + i1)K1(1) + (-1 – i1)K2(1) dt 1 = (-1 + i1)K1 + (-1 – i1)K2 K1 = -iK2 – 0.5 – i0.5 2 = -iK2 – 0.5 – i0.5 + K2 2 + 0.5 + i0.5 = -iK2 + K2 2.5 + i0.5 = [-i + 1]K2 K2 = 1 + i1.5 2 = K1 + 1 + i1.5 1 = K1 + i1.5 K1 = 1 – i1.5 v(t) = (1 – i1.5)e(-1 + i1)t + (1 + i1.5)e(-1 – i1)t (b) d2v

dv

+2 + 4v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 4v = 0 s2 + 2s + 4 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2 ± √22 – 4(1)(4)

1 2 4

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} s1, s2 = s1, s2 =

2(1) -2 ± √4 – 16 2(1)

s1, s2 = s1, s2 =

-2 ± √-12 2(1) -2 ± i3.464

2 s1, s2 = (-2 + i3.464)/2, (-2 – i3.464)/2 s1, s2 = (-1 + i1.732), (-1 – i1.732) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1.732)t + K2e(-1 – i1.732)t … (i) At t = 0+ v(0+) = K1e(-1 + i1.732)0+ + K2e(-1 – i1.732)0+ 2 = K1e0 + K2e0 2 = K1(1) + K2(1) 2 = K1 + K2 Differentiating (i) with respect to ‘t’ dv(t) = (-1 + i1.732)K1e(-1 + i1.732)t + (-1 – i1.732)K2e(-1 – i1.732)t dt dv(0+) = (-1 + i1.732)K1e(-1 + i1.732)0+ + (-1 – i1.732)K2e(-1 – i1.732)0+ dt 1 = (-1 + i1.732)K1e0 + (-1 – i1.732)K2e0 1 = (-1 + i1.732)K1(1) + (-1 – i1.732)K2(1) 1 = (-1 + i1.732)K1 + (-1 – i1.732)K2 (-1 + i1.732)K1 = (1 + i1.732)K2 + 1 K1 = [0.5 – i0.866]K2 - 0.25 - i0.433 2 = [0.5 – i0.866]K2 - 0.25 - i0.433 + K2 2 + 0.25 + i0.433 = [0.5 – i0.866]K2 + K2 2.25 + i0.433 = [0.5 – i0.866 + 1]K2 2.25 + i0.433 = [1.5 – i0.866]K2 K2 = 1 + i0.867 2 = K1 + 1 + i0.867

51

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

52

1 – i0.867 = K1 Putting corresponding values v(t) = (1 – i0.867)e(-1 + i1.732)t + (1 + i0.867)e(-1 – i1.732)t (c) d2v

dv +4 + 2v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 4sv + 2v = 0 s2 + 4s + 2 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-4 ± √42 – 4(1)(2) 2(1) -4 ± √16 – 8 2(1) -4 ± √8 2(1) -4 ± 2.828

2 s1, s2 = (-4 + 2.828)/2, (-4 – 2.828)/2 s1, s2 = (-0.586), (-3.414) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-0.586)t + K2e(-3.414)t … (i)

1 4 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

53

At t = 0+ v(0+) = K1e(-0.586)0+ + K2e(-3.414)0+ 2 = K1e0 + K2e0 2 = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-0.586)K1e(-0.586)t + (-3.414)K2e(-3.414)t dt dv(0+) = (-0.586)K1e(-0.586)0+ + (-3.414)K2e(-3.414)0+ dt 1 = (-0.586)K1e0 + (-3.414)K2e0 1 = (-0.586)K1(1) + (-3.414)K2(1) 1 = -0.586K1 - 3.414K2 K1 = -1.706 – 5.826K2 2 = [-1.706 – 5.826K2] + K2 2 = -1.706 – 5.826K2 + K2 2 + 1.706 = [–5.826 + 1]K2 3.706 = [–4.826]K2 K2 = -0.768 2 = K1 + (-0.768) 2.768 = K1 Putting corresponding values we get v(t) = 2.768e(-0.586)t – 0.768e(-3.414)t (d) d2v dv 2 +8 + 16v = 0 dt2 dt After simplification d2v

dv +4 + 8v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 4sv + 8v = 0 s2 + 4s + 8 = 0 a b c

1 4 8

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-4 ± √42 – 4(1)(8) 2(1) -4 ± √16 – 32 2(1) -4 ± √-16 2(1) -4 ± i4

2 s1, s2 = (-4 + i4)/2, (-4 – i4)/2 s1, s2 = (-2 + i2), (-2 – i2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-2 + i2)t + K2e(-2 – i2)t At t = 0+ v(0+) = K1e(-2 + i2)0+ + K2e(-2 – i2)0+ 2 = K1e0 + K2e0 2 = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-2 + i2)K1e(-2 + i2)t + (-2 – i2)K2e(-2 – i2)t dt dv(0+) = (-2 + i2)K1e(-2 + i2)0+ + (-2 – i2)K2e(-2 – i2)0+ dt 1 = (-2 + i2)K1e0 + (-2 – i2)K2e0 1 = (-2 + i2)K1(1) + (-2 – i2)K2(1) 1 = (-2 + i2)K1 + (-2 – i2)K2 K1 = -i – 0.25 – i0.25 K1 = -i1.25 – 0.25

54

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

55

2 = -i1.25 – 0.25 + K2 2 + i1.25 + 0.25 = K2 2.25 + i1.25 = K2 v(t) = [-i1.25 – 0.25]e(-2 + i2)t + [2.25 + i1.25]e(-2 – i2)t (e) d2v

dv +2 + 3v = 0 2 dt dt Characteristic equation: as2 + bs + c = 0 Here s2v + 2sv + 3v = 0 s2 + 2s + 3 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 =

-2 ± √22 – 4(1)(3) 2(1) -2 ± √4 – 12 2(1) -2 ± √-8

s1, s2 =

2(1) -2 ± i2.828

2 s1, s2 = (-2 + i2.828)/2, (-2 – i2.828)/2 s1, s2 = (-1 + i1.414), (-1 – i1.414) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1 + i1.414)t + K2e(-1 – i1.414)t At t = 0+

1 2 3

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

56

v(0+) = K1e(-1 + i1.414)0+ + K2e(-1 – i1.414)0+ 2 = K1e0 + K2e0 2 = K1(1) + K2(1) 2 = K1 + K2 Differentiating (i) with respect to ‘t’ dv(t) = (-1 + i1.414)K1e(-1 + i1.414)t + (-1 – i1.414)K2e(-1 – i1.414)t dt dv(0+) = (-1 + i1.414)K1e(-1 + i1.414)0+ + (-1 – i1.414)K2e(-1 – i1.414)0+ dt dv(0+) = (-1 + i1.414)K1e0 + (-1 – i1.414)K2e0 dt dv(0+) = (-1 + i1.414)K1(1) + (-1 – i1.414)K2(1) dt 1 = (-1 + i1.414)K1 + (-1 – i1.414)K2 K1 = [0.334 – i0.943]K2 – 0.334 – i0.471 2 = [0.334 – i0.943]K2 – 0.334 – i0.471 + K2 2 + 0.334 + i0.471 = [1.334 – i0.943]K2 2.334 + i0.471 = [1.334 – i0.943]K2 K2 = 1 + i1.060 2 = K1 + 1 + i1.060 1 - i1.060 = K1 Putting corresponding values we get v(t) = (1 - i1.060)e(-1 + i1.414)t + (1 + i1.060)e(-1 – i1.414)t (f) d2v

dv

+3 + 5v = 0 dt2 dt Characteristic equation: as2 + bs + c = 0 Here s2v + 3sv + 5v = 0 s2 + 3s + 5 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

1 3 5

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 2a Putting corresponding values we get s1, s2 =

-3 ± √32 – 4(1)(5) 2(1)

s1, s2 = s1, s2 = s1, s2 =

-3 ± √9 – 20 2(1) -3 ± √-11 2(1) -3 ± i3.317

2 s1, s2 = (-3 + i3.317)/2, (-3 – i3.317)/2 s1, s2 = (-1.5 + i1.659), (-1.5 – i1.659) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get v(t) = K1e(-1.5 + i1.659)t + K2e(-1.5 – i1.659)t … (i) At t = 0+ v(0+) = K1e(-1.5 + i1.659)0+ + K2e(-1.5 – i1.659)0+ 2 = K1e0 + K2e0 2 = K1(1) + K2(1) 2 = K1 + K2 … (ii) Differentiating (i) with respect to ‘t’ dv(t) = (-1.5 + i1.659)K1e(-1.5 + i1.659)t + (-1.5 – i1.659)K2e(-1.5 – i1.659)t dt dv(0+) = (-1.5 + i1.659)K1e(-1.5 + i1.659)0+ + (-1.5 – i1.659)K2e(-1.5 – i1.659)0+ dt dv(0+) = (-1.5 + i1.659)K1e0 + (-1.5 – i1.659)K2e0 dt dv(0+) = (-1.5 + i1.659)K1(1) + (-1.5 – i1.659)K2(1) dt 1 = (-1.5 + i1.659)K1 + (-1.5 – i1.659)K2 K1 = 0.101 – i0.996 – 0.300 – i0.332

57

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

58

K1 = -0.199 – i1.328 2 = -0.199 – i1.328 + K2 2 + 0.199 + i1.328 = K2 2.199 + i1.328 = K2 Putting corresponding values we get v(t) = (-0.199 – i1.328)e(-1.5 + i1.659)t + (2.199 + i1.328)e(-1.5 – i1.659)t Q#6.9: Solve the differential equation d3i d2i di 3 +8 + 10 + 3i = 0 dt dt dt 3s3i + 8s2i + 10si + 3i = 0 X-TICS EQUATION: 3s3 + 8s2 + 10s + 3 = 0 We can find roots using synthetic division 1 -1 1

8

10

3

-1

-07

-3

3

0

7

s = -1 So s + 1 is a factor. (s + 1)(s2 + 7s + 3) = 0 s2 + 7s + 3 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -7 ± √72 – 4(1)(3) s1, s2 = 2(1) -7 ± √49 – 12 s1, s2 = 2(1) -7 ± √37 s1, s2 = 2(1) -7 ± 6.083 s1, s2 =

Remainder is zero.

1 7 3

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

59

2 s1, s2 = (-7 + 6.083)/2, (-7 – 6.083)/2 s1, s2 = (-0.917)/2, (-13.083)/2 s1, s2 = -0.459, -6.542 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-0.459t + K2e-6.542t So –1, -0.459, -6.542 are roots of the X-TICS equation. General solution: i(t) = K1e-1t + K2e-0.459t + K3e-6.542t Q#6.10: Solve the differential equation d3i d2i di 2 +9 + 13 + 6i = 0 dt dt dt 2s3i + 9s2i + 13si + 6i = 0 X-TICS EQUATION: 2s3 + 9s2 + 13s + 6 = 0 We can find roots using synthetic division 2 -1 2

9

13

6

-2

-07

-6

6

0

7

s = -1 So s + 1 is a factor. (s + 1)(2s2 + 7s + 6) = 0 2s2 + 7s + 6 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -7 ± √72 – 4(1)(6) s1, s2 = 2(1) -7 ± √49 – 24 s1, s2 =

Remainder is zero.

2 7 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

2(1) -7 ± √25 2(1)

s1, s2 =

-7 ± 5

2 s1, s2 = (-7 + 5)/2, (-7 – 5)/2 s1, s2 = (-2)/2, (-12)/2 s1, s2 = -1, -6 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-1t + K2e-6t So –1, -1, -6 are roots of the X-TICS equation. General solution: i(t) = K1e-1t + K2e-1t + K3e-6t … (i) At t = 0+ i(0+) = K1e-1(0+) + K2e-1(0+) + K3e-6(0+) 0 = K1e0 + K2e0 + K3e0 0 = K1(1) + K2(1) + K3(1) 0 = K1 + K2 + K3 Differentiating (i) with respect to ‘t’ di = -K1e-t - K2e-1t - 6K3e-6t dt di(0+) = -K1e-(0+) - K2e-1(0+) - 6K3e-6(0+) dt 1 = -K1e0 - K2e0 - 6K3e0 1 = -K1(1) - K2(1) - 6K3(1) 1 = -K1 - K2 - 6K3 d2i = K1e-t + K2e-1t + 6K3e-6t 2

dt d2i(0+) = K1e-(0+) + K2e-1(0+) + 6K3e-6(0+) dt2

60

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

61

-1 = K1e0 + K2e0 + 6K3e0 -1 = K1(1) + K2(1) + 6K3(1) -1 = K1 + K2 + 6K3 Q#6.11: The response of a network is found to be α i = K1te- t, t ≥ 0 where α is real and positive. Find the time at which i(t) attains a maximum value. K=1 α =4 t=0 i(0) = 0 t=1 i(1) = 0.018 t=2 i(2) = 0.001 t=3 i(3) = 0

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

62

3.5

3

3

2.5

2

2

current

Series1 Series2

1.5

1

1

0.5

0

1

0

0.018

0.001

2

3 2

3

0.001

0

Series1

0

1

Series2

0

0.018 tim e

4

0

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

63

Q#6.12: In a certain network, it is found that the current is given by the expression α α i = K1e- 1t – K2e- 2t, t > 0, α 1 > α 2 Show that i(t) reaches a maximum value at time 1 α 1K1 t= ln α 1-α 2 α 2K2 1 α 1K1 t= ln α 1 K2 Solution: α α i(t) = K1e- 1t – K2e- 2t α α i(1) = K1e- 1(1) – K2e- 2(1) t=1 α α i(1) = K1e- 1 – K2e- 2 α α i(2) = K1e- 1(2) – K2e- 2(2) t=2 α α i(2) = K1e-2 1 – K2e-2 2 α α i(3) = K1e- 1(3) – K2e- 2(3) t=3 α α i(3) = K1e-3 1 – K2e-3 2 α α α i((1/α 1)ln(α 1K1/K2)) = K1e- 1((1/ 1)ln( 1K1/K2)) α α α t = (1/α 1)ln(α 1K1/K2) – K2e- 2((1/ 1)ln( 1K1/K2)) 1 t=

α 1K1

ln α 1 K2 Q#6.13: The graph shows a damped sinusoidal waveform having the general form σ Ke- t sin(ω t + φ ) From the graph, determine numerical values for K, σ , ω , and φ . t = 1 sec 1 f= t f = 1 Hz ω = 2π f ω = 2π (1) ω = 2π ω = 6.28 rad/sec t V(t) 0 0 0.4 0 0.8 0 Q#6.14: Repeat Prob. 6.13 for the waveform of the accompanying figure. t = 4 msec 1 f= t

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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f = 250 Hz ω = 2π f ω = 2π (250) ω = 1570 rad/sec φ = 00 Q#6.15: In the network of the figure, the switch K is closed and a steady state is reached in the network. At t = 0, the switch is opened. Find an expression for the current in the inductor, i2(t). Solution: Circuit diagram:

K +

10 Ω

100 V -

Equivalent circuit before switching:

At t = 0100 i2(0-) = 10 i2(0-) = 10 A = i2(0+) VC(0-) = VC(0+) = 0 Equivalent circuit after switching:

i2 1H 20 µ F

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

i2

Using KVL we have di2 1 L + ∫ i2dt = 0 … (i) dt C Differentiating we have d2i2 i2 L + =0 2 dt C The characteristic equation is i2 Ls2i2 + = 0 C 1 Ls2 + =0 C 1 2 s =LC -1 s=±

LC 1

s=± j

LC

Roots are pure imaginary so from table 6-1 Form of solution i = K1cos ω t + K2 sin ω t Here

65

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 1 ω =

LC

i = i2(t) i2(t) = K1cos (1/LC)1/2t + K2 sin (1/LC)1/2t … (ii) From equation (i) at t = 0+ di2(0+) L + VC(0+) = 0 dt Putting corresponding values di2(0+) = 0 Amp./sec dt From (ii) at t = 0+ i2(0+) = K1cos (1/LC)1/2(0+) + K2 sin (1/LC)1/2(0+) i2(0+) = K1(1) + 0 K1 = 10 di2(t) = -(1/LC)1/2K1sin (1/LC)1/2t + (1/LC)1/2K2 cos (1/LC)1/2t dt At t = 0+ di2(0+) = -(1/LC)1/2K1sin (1/LC)1/2(0+) + (1/LC)1/2K2 cos (1/LC)1/2(0+) dt di2(0+) = -(1/LC)1/2K1sin (1/LC)1/2(0+) + (1/LC)1/2K2 cos (1/LC)1/2(0+) dt 0 = (1/LC)1/2K2 0 = K2 i2(t) = (10)cos (1/LC)1/2(t) + (0) sin (1/LC)1/2(t) i2(t) = (10)cos (1/(1)(20(10-6)))1/2(t) i2(t) = 10cos 223.607t Q#6.16: The capacitor of the figure has an initial voltage VC(0-) = V1, and at the same time the current in the inductor is zero. At t = 0, the switch K is closed. Determine an expression for the voltage V2(t). Solution:

66

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

V1 C

V2 R

Here VC(0-) = VC(0+) = 0 = V2(0+) iL(0-) = iL(0+) = 0 After switching dV2 V2 1 C + + ∫ V2dt = 0 … (i) dt R L Differentiating with respect to ‘t’ d2V2 dV2 V2 C + + =0 dt2 Rdt L d2V2 dV2 V2 + + =0 2 dt RCdt LC The characteristic equation is s V2 2 s V2 + V2 + =0 RC LC s 1 s2 + + =0 RC LC a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 =

67

-(1/RC) ± √(1/RC)2 – 4(1)(1/LC) 2(1) -(1/RC) ± √(1/R2C2) – (4/LC)

L

1 1/RC 1/LC

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} s1, s2 = 2 s1, s2 = -(1/2RC) ± √(1/2RC)2 – (1/LC) Case (a) if (1/2RC)2 – (1/LC) = 0 (1/2RC)2 = (1/LC) that is 1 L R= 2 C Then 1 s=2RC So form of solution is: Nature of roots s1, s2 = (-1/2RC + 0), (-1/2RC - 0) s1, s2 = (-1/2RC), (-1/2RC) Negative real and equal V2(t) = K1es1t + K2tes2t V2(t) = K1e(-1/2RC)t + K2te(-1/2RC)t … (ii) From equation (i) at t = 0+ dV2(0+) V2(0+) C + + iL(0+) = 0 dt R Putting corresponding values dV2(0+) -V1 = dt RC At t = 0+ V2(0+) = K1e(-1/2RC)0+ + K2(0+)e(-1/2RC)0+ … (ii) V1 = K1e0 V1 = K1 Differentiating (ii) with respect to ‘t’ dV2(t) = (-1/2RC)K1e(-1/2RC)t + K2[-te(-1/2RC)t - e(-1/2RC)t] dt At t = 0+ dV2(0+) = (-1/2RC)K1e(-1/2RC)0+ + K2[-(0+)e(-1/2RC)0+ - e(-1/2RC)0+] dt dV2(0+) = (-1/2RC)K1e0 + K2[-e0]

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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dt -V1 = (-1/2RC)K1(1) + K2[-(1)] RC V1 = (1/2RC)K1 + K2 RC Putting corresponding values V1 = (1/2RC)V1 + K2 RC -V1 K2 = 2RC Putting corresponding values V1 V2(t) = V1e

(-1/2RC)t

te(-1/2RC)t

2RC

Q#6.17: The voltage source in the network of the figure is described by the equation, V1 = 2 cos 2t for t ≥ 0 and is a short circuit prior to that time. Determine V2(t). Repeat if V1 = K1t for t ≥ 0 and V1 = 0 for t < 0. Solution:

+ -

V1

Here iL(0-) = iL(0+) = 0 A VC(0-) = VC(0+) = 0 = V2(0+) For t ≥ 0 According to KCL V2 1 dV2 1 + + ∫ (V2 – V1)dt = 0 … (i) 2 2 dt 1 Differentiating we have 1 d2V2

dV2 + 2dt

+ V2 – V1 = 0 2 dt2

V2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

70

d2V2

dV2 +

+ 2V2 – 2V1 = 0 2

dt dV2

dt d2V2 +

+ 2V2 = 2V1 dt2 d2V2

dt dV2 + dt dV2

+ 2V2 = 2[2 cos 2t] 2

dt d2V2

+ + 2V2 = 4 cos 2t … (ii) dt dt2 Equation (ii) is a non-homogeneous differential equation Characteristic equation is s2V2 + sV2 + 2V2 = 0 s2 + s + 2 = 0 a b c

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-1 ± √12 – 4(1)(2) 2(1) -1 ± √1 – 8 2(1) -1 ± √-7 2(1) -1 ± i2.646

2 s1, s2 = (-1 + i2.646)/2, (-1 – i2.646)/2 s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get V2(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 – i1.323)t … (ii) Complementary solution.

1 1 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} The form of particular solution is V2P = Acos 2t + Bsin 2t … (iii) Substituting (iii) in (ii) d2 d [Acos 2t + Bsin 2t] + [Acos 2t + Bsin 2t] + 2[Acos 2t + Bsin 2t] = 4 cos 2t dt2 dt After simplification -2Asin 2t + 2B cos 2t – 4Acos 2t – 4Bsin 2t + 2Acos 2t + 2Bsin 2t = 4 cos 2t Equating coefficients of L.H.S & R.H.S -2A – 4B + 2B = 0 -2A – 2B = 0 -2A = 2B -A = B 2B – 4A + 2A = 4 2B – 2A = 4 B–A=2 -A – A = 2 -2A = 2 A = -1 -(-1) = B B=1 V2P = -1cos 2t + 1sin 2t V2P = -cos 2t + sin 2t Therefore particular solution is V2 = complementary + particular V2 = K1e(-0.5 + i1.323)t + K2e(-0.5 – i1.323)t + [-cos 2t + sin 2t] V2 = K1e(-0.5 + i1.323)t + K2e(-0.5 – i1.323)t - cos 2t + sin 2t … (iV) dV2 = (-0.5 + i1.323)K1e(-0.5 + i1.323)t + (-0.5 – i1.323)K2e(-0.5 – i1.323)t + 2sin 2t + 2cos 2t dt V2(0+) 1 dV2(0+) + + iL(0+) = 0 2 2 dt Putting corresponding values dV2(0+) =0 dt At t = 0+ dV2(0+) = dt (-0.5 + i1.323)K1e(-0.5 + i1.323)0+ + (-0.5 – i1.323)K2e(-0.5 – i1.323)0+ + 2sin 2(0+) + 2cos 2(0+) 0 = (-0.5 + i1.323)K1e0 + (-0.5 – i1.323)K2e0 + 2sin 0 + 2cos 0 0 = (-0.5 + i1.323)K1(1) + (-0.5 – i1.323)K2(1) + 2(1) 0 = (-0.5 + i1.323)K1 + (-0.5 – i1.323)K2 + 2 … (V) At t = 0+ equation (iV) V2(0+) = K1e(-0.5 + i1.323)0+ + K2e(-0.5 – i1.323)0+ - cos 2(0+) + sin 2(0+)

71

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

72

0 = K1e0 + K2e0 – 1 0 = K1(1) + K2(1) – 1 0 = K1 + K2 – 1 1 = K1 + K2 … (Vi) K1 = 1 – K2 Putting the value of K1 in (V) 0 = (-0.5 + i1.323)(1 – K2) + (-0.5 – i1.323)K2 + 2 0 = (-0.5 + i1.323) – K2(-0.5 + i1.323) + (-0.5 – i1.323)K2 + 2 0 = -0.5 + i1.323 + 0.5K2 – i1.323K2 – 0.5K2 - i1.323K2 + 2 -1.5 - i1.323 = [0.5 – i1.323 – 0.5 - i1.323]K2 -1.5 - i1.323 = [–i2.646]K2 1.5 + i1.323 = [i2.646]K2 K2 = 0.5 – i0.567 Putting the value of K2 in (Vi) 1 = K1 + 0.5 – i0.567 0.5 + i0.567 = K1 V2 = [0.5 + i0.567]e(-0.5 + i1.323)t + [0.5 – i0.567]e(-0.5 – i1.323)t - cos 2t + sin 2t Part (b) If V1 = Kt dV2 d2V2 + + 2V2 = 2V1 dt dt2 dV2 d2V2 + + 2V2 = 2K1 … (a) 2 dt dt Equation (ii) is a non-homogeneous differential equation Characteristic equation is s2V2 + sV2 + 2V2 = 0 s2 + s + 2 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -1 ± √12 – 4(1)(2)

1 1 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} s1, s2 = s1, s2 = s1, s2 = s1, s2 =

2(1) -1 ± √1 – 8 2(1) -1 ± √-7 2(1) -1 ± i2.646

2 s1, s2 = (-1 + i2.646)/2, (-1 – i2.646)/2 s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323) General solution: v(t) = C1es1t + C2es2t Putting corresponding values we get V2(t) = C1e(-0.5 + i1.323)t + C2e(-0.5 – i1.323)t … (i) Complementary solution. The form of particular solution is From table 6.2 A1tn B0tn + B1tn- 1 + … + Bn – 1t + Bn V2P = At + B … (ii) Substituting equation (ii) in (a) d2 d [At + B] + [At + B] + 2[At + B] = 2K1t 2 dt dt After simplification A + 2At + 2B = 2K1t Equating coefficients t: 2A = 2K1 A = K1 Constants: A + 2B = 0 K1 + 2B = 0 B = -0.5K1 V2P = K1t - 0.5K1 Therefore Particular solution = Complementary + particular V2 = C1e(-0.5 + i1.323)t + C2e(-0.5 – i1.323)t + K1t - 0.5K1 … (ii) Differentiating dV2

73

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

74

= (-0.5 + i1.323)C1e(-0.5 + i1.323)t + (-0.5 – i1.323)C2e(-0.5 – i1.323)t + K1 dt At t = 0+ eq. (ii) V2(0+) = C1e(-0.5 + i1.323)0+ + C2e(-0.5 – i1.323)0+ + K1(0+) - 0.5K1 0 = C1e0 + C2e0 - 0.5K1 0 = C1(1) + C2(1) - 0.5K1 0 = C1 + C2 - 0.5K1 … (iii) dV2(0+) = (-0.5 + i1.323)C1e(-0.5 + i1.323)0+ + (-0.5 – i1.323)C2e(-0.5 – i1.323)0+ + K1 dt 0 = (-0.5 + i1.323)C1e0 + (-0.5 – i1.323)C2e0 + K1 0 = (-0.5 + i1.323)C1(1) + (-0.5 – i1.323)C2(1) + K1 0 = (-0.5 + i1.323)C1 + (-0.5 – i1.323)C2 + K1 … (iV) Putting the value of C1 from (iii) into (iV) 0 = (-0.5 + i1.323)[0.5K1 – C2] + (-0.5 – i1.323)C2 + K1 After simplification C2 = 0.250 – i0.284K1 Putting the value of C2 in (iii) 0 = C1 + 0.250 – i0.284K1 - 0.5K1 C1 = -0.250 + i0.284K1 + 0.5K1 Putting corresponding values eq. (ii) becomes V2 = [-0.250 + i0.284K1 + 0.5K1]e(-0.5 + i1.323)t + [0.250 – i0.284K1]e(-0.5 – i1.323)t + K1t - 0.5K1 Q#6.18: Solve the following nonhomogeneous differential equations for t ≥ 0. (a) d2i di +2 +i=1 dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 2si + 1i = 0 s2 + 2s + 1 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

1 2 1

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

75

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-2 ± √22 – 4(1)(1) 2(1) -2 ± √4 – 4 2(1) -2 ± √0 2(1) -2 ± 0

2 s1, s2 = (-2 + 0)/2, (-2 – 0)/2 s1, s2 = -1, -1 General solution: i(t) = K1es1t + K2tes2t Putting corresponding values we get i(t) = K1e(-1)t + K2te(-1)t Particular solution From table 6.2 V(a constant) iP = A = 1

A

i(t) = K1e(-1)t + K2te(-1)t + 1 (b) d2i

di + 3 + 2i = 5t …(c) dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 A B C 2 -b ± √b – 4ac s1, s2 = 2a

1 3 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 =

-3 ± √32 – 4(1)(2) 2(1) -3 ± √9 – 8 2(1) -3 ± √1 2(1) -3 ± 1

s1, s2 =

2 s1, s2 = (-3 + 1)/2, (-3 – 1)/2 s1, s2 = (-1), (-2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2e(-2)t … (i) Particular solution From table 6.2 A1tn B0tn + B1tn- 1 + … + Bn – 1t + Bn iP = At + B … (ii) Putting the value of iP from (ii) into eq. (c) d2 d [At + B] + 3 [At + B] + 2[At + B] = 5t dt2 dt After simplification 3A + 2At + 2B = 5t Equating coefficients 2A = 5 A = 2.5 Constants 3A + 2B = 0 3(2.5) + 2B = 0 7.5 + 2B = 0 B = -3.75 iP = 2.5t – 3.75 iP = -15/4 + 5t/2 i(t) = K1e(-1)t + K2e(-2)t - 15/4 + 5t/2 (c) d2i

di

76

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} + 3 + 2i = 10 sin 10t … (k) dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 A B C -b ± √b2 – 4ac s1, s2 = 2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

77

1 3 2

-3 ± √32 – 4(1)(2) 2(1) -3 ± √9 – 8 2(1) -3 ± √1 2(1) -3 ± 1

2 s1, s2 = (-3 + 1)/2, (-3 – 1)/2 s1, s2 = (-1), (-2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2e(-2)t … (i) Particular solution From table 6.2 iP = A sin 10t + B cos 10t Putting the value of iP in (k) d2 d [A sin 10t + B cos 10t] + 3 [A sin 10t + B cos 10t] + 2[A sin 10t + B cos 10t] = dt2 dt 10 sin 10t -100A sin 10t – 100B cos 10t + 3[10A cos 10t - 10B sin 10t] + 2A sin 10t + 2B cos 10t = 10 sin 10t

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

78

-100A sin 10t – 100B cos 10t + 30A cos 10t - 30B sin 10t + 2A sin 10t + 2B cos 10t = 10 sin 10t Equating coefficients Sin -100A – 30B + 2A = 10 -98A – 30B = 10 Cos -100B + 30A + 2B = 0 -98B + 30A = 0 B = 0.306A -98A – 30(0.306A) = 10 -98A – 9.18A = 10 -107.18A = 10 A = -0.094 B = 0.306(-0.094) B = -0.029 Now iP = -0.094 sin 10t + (-0.029) cos 10t iP = -0.094 sin 10t - 0.029 cos 10t i(t) = K1e(-1)t + K2e(-2)t + [-0.094 sin 10t - 0.029 cos 10t] i(t) = K1e(-1)t + K2e(-2)t - 0.094 sin 10t - 0.029 cos 10t (d) d2q

dq +5 + 6q = te-t … (m) dt2 dt Nonhomogeneous differential equation q(t) = qC + qP Complementary solution Characteristic equation is s2q + 5sq + 6q = 0 s2 + 5s + 6 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 =

-5 ± √52 – 4(1)(6) 2(1)

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 = s1, s2 = s1, s2 =

-5 ± √25 – 24 2(1) -5 ± √1 2(1) -5 ± 1

2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: q(t) = K1es1t + K2es2t Putting corresponding values we get q(t) = K1e-2t + K2e-3t Particular solution qP = (At + B)te-t ⇒ multiplied by ‘t’ because e-t is repeating in both iP & iC Putting the value of iP in eq. m d2 d [(At + B)te-t] + 5 [(At + B)te-t] + 6[(At + B)te-t] = te-t dt2 dt At2e-t - 2Ate-t – 2Ate-t + 2Ae-t + Bte-t – Be-t – Be-t + 5[-At2e-t + 2Ate-t - Bte-t + Be-t] + 6[At2e-t + Bte-t] = te-t At2e-t - 2Ate-t – 2Ate-t + 2Ae-t - Bte-t – Be-t – Be-t - 5At2e-t + 10Ate-t - 5Bte-t + 5Be-t + 6At2e-t + 6Bte-t = te-t Equating coefficients t2e-t: A – 5A + 6A = 0 2A = 0 A=0 te-t: -2A – 2A + 10A = 1 6B = 3 B = 0.5 e-t: 2 – B – B + 5B = 0 2 + 3B = 0 B = -0.667 When B = 0.5 qP = (0t + (0.5))te-t qP = 0.5te-t When B = -0.667

79

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

80

qP = (0t + (-0.667))te-t qP = -0.667te-t q(t) = K1e-2t + K2e-3t - 0.667te-t, K1e-2t + K2e-3t + 0.5te-t (e) d2V

dV

+5 + 6V = e-2t + 5e-3t … (n) dt dt Nonhomogeneous differential equation V(t) = VC + VP Complementary solution Characteristic equation is s2V + 5sV + 6V = 0 s2 + 5s + 6 = 0 A B C 2

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -5 ± √52 – 4(1)(6) s1, s2 = 2(1) -5 ± √25 – 24 s1, s2 = 2(1) -5 ± √1 s1, s2 = 2(1) -5 ± 1 s1, s2 = 2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: V(t) = K1es1t + K2es2t Putting corresponding values we get V(t) = K1e-2t + K2e-3t Particular solution VP = Ce-2t + Ce-3t Substituting the value of VP in eq. (n)

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

81

d2

d [Ce-2t + Ce-3t] + 5 [Ce-2t + Ce-3t] + 6[Ce-2t + Ce-3t] = e-2t + 5e-3t dt2 dt 4Ce-2t + 9Ce-3t + 5[-2Ce-2t – 3Ce-3t] + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t 4Ce-2t + 9Ce-3t - 10Ce-2t – 15Ce-3t + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t Equating coefficients e-2t 4C – 10C + 6C = 1 4C – 10C + 6C = 1 C=0 VP = (0)e-2t + (0)e-3t VP = 0 V(t) = K1e-2t + K2e-3t + 0 V(t) = K1e-2t + K2e-3t Q#6.19: Solve the differential equations given in Prob. 6.18 subject to the following initial conditions: dx x(0+) = 1 & (0+) = -1 dt where x is the general dependent variable. (a) d2i di +2 +i=1 dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 2si + 1i = 0 s2 + 2s + 1 = 0 A 1 B 2 C 1

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2 ± √22 – 4(1)(1) s1, s2 = 2(1) -2 ± √4 – 4 s1, s2 = 2(1) -2 ± √0

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

82

s1, s2 = 2(1) -2 ± 0

s1, s2 =

2 s1, s2 = (-2 + 0)/2, (-2 – 0)/2 s1, s2 = -1, -1 General solution: i(t) = K1es1t + K2tes2t Putting corresponding values we get i(t) = K1e(-1)t + K2te(-1)t Particular solution From table 6.2 V(a constant) iP = A = 1 i(t) = K1e(-1)t + K2te(-1)t + 1 At t = 0+ i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+ + 1 1 = K1e0 + 1 1 = K1(1) + 1 K1 = 0 di (0+) = -K1e-(0+) - K2(0+)e-(0+) + K2e-(0+) dt -1 = -K1e0 + K2e0 -1 = -K1(1) + K2(1) -1 = -K1 + K2 -1 = -(0) + K2 -1 = K2 i(t) = (0)e(-1)t + (-1)te(-1)t + 1 i(t) = - te-t + 1 (b) d2i

di

+ 3 + 2i = 5t …(c) dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 3si + 2i = 0 s2 + 3s + 2 = 0

A

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

A B C 2 -b ± √b – 4ac

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-3 ± √32 – 4(1)(2) 2(1) -3 ± √9 – 8 2(1) -3 ± √1 2(1) -3 ± 1

2 s1, s2 = (-3 + 1)/2, (-3 – 1)/2 s1, s2 = (-1), (-2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2e(-2)t … (i) Particular solution From table 6.2 A1tn B0tn + B1tn- 1 + … + Bn – 1t + Bn iP = At + B … (ii) Putting the value of iP from (ii) into eq. (c) d2 d [At + B] + 3 [At + B] + 2[At + B] = 5t dt2 dt After simplification 3A + 2At + 2B = 5t Equating coefficients 2A = 5 A = 2.5 Constants 3A + 2B = 0 3(2.5) + 2B = 0 7.5 + 2B = 0 B = -3.75

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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iP = 2.5t – 3.75 iP = -15/4 + 5t/2 i(t) = K1e(-1)t + K2e(-2)t - 15/4 + 5t/2 … (b) At t = 0+ i(0+) = K1e(-1)0+ + K2e(-2)0+ - 15/4 + 5(0+)/2 1 = K1e0 + K2e0 - 15/4 1 = K1(1) + K2(1) - 15/4 1 = K1 + K2 - 15/4 1 + 15/4 = K1 + K2 4.75 = K1 + K2 … (A) Differentiating (b) at t = 0+ di (0+) = -K1e(-1)0+ - 2K2e(-2)0+ + 5/2 dt -1 = -K1e0 - 2K2e0 + 5/2 -1 = -K1(1) - 2K2(1) + 5/2 -1 = -K1 - 2K2 + 5/2 -1 – 2.5 = -K1 - 2K2 -3.5 = -K1 - 2K2 From (A) K1 = 4.75 – K2 -3.5 = -[4.75 – K2] - 2K2 -3.5 = -4.75 + K2 - 2K2 -3.5 + 4.75 = -K2 1.25 = -K2 K2 = -1.25 4.75 = K1 + (-1.25) 4.75 + 1.25= K1 K1 = 6 Putting corresponding values of K1 & K2 i(t) = 6e(-1)t + (-1.25)e(-2)t - 15/4 + 5t/2 i(t) = 6e(-1)t - 1.25e(-2)t - 15/4 + 5t/2 (c) d2i

di + 3 + 2i = 10 sin 10t … (k) dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 A B

1 3

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} C -b ± √b – 4ac

85 2

2

s1, s2 =

2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-3 ± √32 – 4(1)(2) 2(1) -3 ± √9 – 8 2(1) -3 ± √1 2(1) -3 ± 1

2 s1, s2 = (-3 + 1)/2, (-3 – 1)/2 s1, s2 = (-1), (-2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2e(-2)t … (i) Particular solution From table 6.2 iP = A sin 10t + B cos 10t Putting the value of iP in (k) d2 d [A sin 10t + B cos 10t] + 3 [A sin 10t + B cos 10t] + 2[A sin 10t + B cos 10t] = dt2 dt 10 sin 10t -100A sin 10t – 100B cos 10t + 3[10A cos 10t - 10B sin 10t] + 2A sin 10t + 2B cos 10t = 10 sin 10t -100A sin 10t – 100B cos 10t + 30A cos 10t - 30B sin 10t + 2A sin 10t + 2B cos 10t = 10 sin 10t Equating coefficients Sin -100A – 30B + 2A = 10 -98A – 30B = 10 Cos -100B + 30A + 2B = 0 -98B + 30A = 0 B = 0.306A

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} -98A – 30(0.306A) = 10 -98A – 9.18A = 10 -107.18A = 10 A = -0.094 B = 0.306(-0.094) B = -0.029 Now iP = -0.094 sin 10t + (-0.029) cos 10t iP = -0.094 sin 10t - 0.029 cos 10t i(t) = K1e(-1)t + K2e(-2)t + [-0.094 sin 10t - 0.029 cos 10t] i(t) = K1e(-1)t + K2e(-2)t - 0.094 sin 10t - 0.029 cos 10t At t = 0+ i(0+) = K1e(-1)0+ + K2e(-2)0+ - 0.094 sin 10(0+) - 0.029 cos 10(0+) … (c) 1 = K1e0 + K2e0 - 0.094 sin 0 - 0.029 cos 0 1 = K1(1) + K2(1) - 0.029 (1) 1 = K1 + K2 - 0.029 1.029 = K1 + K2 … (y) Differentiating eq. (c) with respect to ‘t’ di (0+) = -K1e-(0+) - 2K2e(-2)0+ - 0.094(10) cos 10(0+) + 0.029(10) sin 10(0+) dt -1 = -K1e0 - 2K2e0 - 0.94 cos 0 + 0.029(10) sin 0 -1 = -K1(1) - 2K2(1) - 0.94 (1) -1 = -K1 - 2K2 - 0.94 0.06 = K1 + 2K2 From eq. (y) K1 = 1.029 – K2 0.06 = [1.029 – K2] + 2K2 0.06 = 1.029 – K2 + 2K2 -0.969 = K2 Putting the value of K2 in (y) 1.029 = K1 + (-0.969) 1.029 + 0.969 = K1 1.998 = K1 i(t) = 1.998e(-1)t + (-0.969)e(-2)t - 0.094 sin 10t - 0.029 cos 10t i(t) = 1.998e(-1)t - 0.969e(-2)t - 0.094 sin 10t - 0.029 cos 10t (d) d2q dq +5 + 6q = te-t … (m) dt2 dt Nonhomogeneous differential equation q(t) = qC + qP Complementary solution Characteristic equation is

86

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

87

s2q + 5sq + 6q = 0 s2 + 5s + 6 = 0 A B C

s1, s2 =

1 5 6

-b ± √b2 – 4ac

2a Putting corresponding values we get -5 ± √52 – 4(1)(6) s1, s2 = 2(1) -5 ± √25 – 24 s1, s2 = 2(1) -5 ± √1 s1, s2 = 2(1) -5 ± 1 s1, s2 = 2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: q(t) = K1es1t + K2es2t Putting corresponding values we get q(t) = K1e-2t + K2e-3t Particular solution qP = (At + B)te-t ⇒ multiplied by ‘t’ because e-t is repeating in both iP & iC Putting the value of iP in eq. m d2 d [(At + B)te-t] + 5 [(At + B)te-t] + 6[(At + B)te-t] = te-t dt2 dt At2e-t - 2Ate-t – 2Ate-t + 2Ae-t + Bte-t – Be-t – Be-t + 5[-At2e-t + 2Ate-t - Bte-t + Be-t] + 6[At2e-t + Bte-t] = te-t At2e-t - 2Ate-t – 2Ate-t + 2Ae-t - Bte-t – Be-t – Be-t - 5At2e-t + 10Ate-t - 5Bte-t + 5Be-t + 6At2e-t + 6Bte-t = te-t Equating coefficients t2e-t: A – 5A + 6A = 0 2A = 0 A=0

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} te-t: -2A – 2A + 10A = 1 6B = 3 B = 0.5 e-t: 2 – B – B + 5B = 0 2 + 3B = 0 B = -0.667 When B = 0.5 qP = (0t + (0.5))te-t qP = 0.5te-t When B = -0.667 qP = (0t + (-0.667))te-t qP = -0.667te-t q(t) = K1e-2t + K2e-3t - 0.667te-t, K1e-2t + K2e-3t + 0.5te-t At t = 0+ q(0+) = K1e-2(0+) + K2e-3(0+) - 0.667(0+)e-(0+), K1e-2(0+) + K2e-3(0+) + 0.5(0+)e-(0+) 1 = K1e0 + K2e0, K1e0 + K2e0 1 = K1(1) + K2(1), K1(1) + K2(1) 1 = K1 + K2, K1 + K2 … (y) q(t) = K1e-2t + K2e-3t - 0.667te-t Differentiating with respect to ‘t’ at t = 0+ dq (0+) = -2K1e-2(0+) -3K2e-3(0+) - 0.667[-(0+)e-(0+) + e-(0+)] dt -1 = -2K1e0 -3K2e0 - 0.667[e0] -1 = -2K1(1) -3K2(1) - 0.667 -1 = -2K1 -3K2 - 0.667 0.333 = 2K1 + 3K2 From eq. (y) K1 = 1 – K2 0.333 = 2[1 – K2] + 3K2 0.333 = 2 – 2K2 + 3K2 0.333 = 2 + K2 K2 = 0.333 - 2 K2 = -1.667 1 = K1 + (-1.667) K1 = 2.667 q(t) = 2.667e-2t + (-1.667)e-3t - 0.667te-t q(t) = 2.667e-2t - 1.667e-3t - 0.667te-t (e) d2V

dV +5 + 6V = e-2t + 5e-3t … (n) dt2 dt Nonhomogeneous differential equation

88

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} V(t) = VC + VP Complementary solution Characteristic equation is s2V + 5sV + 6V = 0 s2 + 5s + 6 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -5 ± √52 – 4(1)(6) s1, s2 = 2(1) -5 ± √25 – 24 s1, s2 = 2(1) -5 ± √1 s1, s2 = 2(1) -5 ± 1 s1, s2 = 2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: V(t) = K1es1t + K2es2t Putting corresponding values we get V(t) = K1e-2t + K2e-3t Particular solution VP = Ce-2t + Ce-3t Substituting the value of VP in eq. (n) d2 d -2t -3t [Ce + Ce ] + 5 [Ce-2t + Ce-3t] + 6[Ce-2t + Ce-3t] = e-2t + 5e-3t dt2 dt 4Ce-2t + 9Ce-3t + 5[-2Ce-2t – 3Ce-3t] + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t 4Ce-2t + 9Ce-3t - 10Ce-2t – 15Ce-3t + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t Equating coefficients e-2t 4C – 10C + 6C = 1 4C – 10C + 6C = 1

89

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

90

C=0 VP = (0)e-2t + (0)e-3t VP = 0 V(t) = K1e-2t + K2e-3t + 0 V(t) = K1e-2t + K2e-3t … (k) At t = 0+ V(0+) = K1e-2(0+) + K2e-3(0+) 1 = K1e0 + K2e0 1 = K1(1) + K2(1) 1 = K1 + K2 … (A) Differentiating eq. (k) with respect to ‘t’ dV (0+) = -2K1e-2(0+) - 3K2e-3(0+) dt -1 = -2K1e0 - 3K2e0 -1 = -2K1(1) - 3K2(1) -1 = -2K1 - 3K2 … (B) K1 = 1 – K2 -1 = -2[1 – K2] - 3K2 … (B) -1 = -2 + 2K2 - 3K2 -1 = K2 1 = K1 + (-1) … (A) 2 = K1 V(t) = 2e-2t + (-1)e-3t … (k) V(t) = 2e-2t - e-3t Q#6.20: Find the particular solutions to the differential equations of Prob. 6.18 for the following initial conditions: dx x(0+) = 2 & (0+) = -1 dt where x is the general dependent variable. (a) d2i di +2 +i=1 dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 2si + 1i = 0 s2 + 2s + 1 = 0 A 1 B 2 C 1 -b ± √b2 – 4ac

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

91

s1, s2 = 2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 =

-2 ± √22 – 4(1)(1) 2(1) -2 ± √4 – 4 2(1) -2 ± √0

s1, s2 =

2(1) -2 ± 0

2 s1, s2 = (-2 + 0)/2, (-2 – 0)/2 s1, s2 = -1, -1 General solution: i(t) = K1es1t + K2tes2t Putting corresponding values we get i(t) = K1e(-1)t + K2te(-1)t Particular solution From table 6.2 V(a constant) iP = A = 1 i(t) = K1e(-1)t + K2te(-1)t + 1 At t = 0+ i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+ + 1 2 = K1e0 + 1 2 = K1(1) + 1 K1 = 1 di (0+) = -K1e-(0+) - K2(0+)e-(0+) + K2e-(0+) dt -1 = -K1e0 + K2e0 -1 = -K1(1) + K2(1) -1 = -K1 + K2 -1 = -(1) + K2 0 = K2 i(t) = (1)e(-1)t + (0)te(-1)t + 1 i(t) = e-t + 1

A

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

92

(b) d2i

di + 3 + 2i = 5t …(c) dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 A B C 2 -b ± √b – 4ac s1, s2 = 2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

-3 ± √32 – 4(1)(2) 2(1) -3 ± √9 – 8 2(1) -3 ± √1 2(1) -3 ± 1

2 s1, s2 = (-3 + 1)/2, (-3 – 1)/2 s1, s2 = (-1), (-2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2e(-2)t … (i) Particular solution From table 6.2 A1tn B0tn + B1tn- 1 + … + Bn – 1t + Bn iP = At + B … (ii) Putting the value of iP from (ii) into eq. (c) d2 d [At + B] + 3 [At + B] + 2[At + B] = 5t

1 3 2

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} dt2 dt After simplification 3A + 2At + 2B = 5t Equating coefficients 2A = 5 A = 2.5 Constants 3A + 2B = 0 3(2.5) + 2B = 0 7.5 + 2B = 0 B = -3.75 iP = 2.5t – 3.75 iP = -15/4 + 5t/2 i(t) = K1e(-1)t + K2e(-2)t - 15/4 + 5t/2 … (b) At t = 0+ i(0+) = K1e(-1)0+ + K2e(-2)0+ - 15/4 + 5(0+)/2 2 = K1e0 + K2e0 - 15/4 2 = K1(1) + K2(1) - 15/4 2 = K1 + K2 - 15/4 2 + 15/4 = K1 + K2 5.75 = K1 + K2 … (A) Differentiating (b) at t = 0+ di (0+) = -K1e(-1)0+ - 2K2e(-2)0+ + 5/2 dt -1 = -K1e0 - 2K2e0 + 5/2 -1 = -K1(1) - 2K2(1) + 5/2 -1 = -K1 - 2K2 + 5/2 -1 – 2.5 = -K1 - 2K2 -3.5 = -K1 - 2K2 From (A) K1 = 5.75 – K2 -3.5 = -[5.75 – K2] - 2K2 -3.5 = -5.75 + K2 - 2K2 -3.5 + 5.75 = -K2 2.25 = -K2 K2 = -2.25 5.75 = K1 + (-2.25) 5.75 + 2.25= K1 K1 = 8 Putting corresponding values of K1 & K2 i(t) = 8e(-1)t + (-2.25)e(-2)t - 15/4 + 5t/2 i(t) = 8e(-1)t - 2.25e(-2)t - 15/4 + 5t/2 (c) d2i

di

93

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} + 3 + 2i = 10 sin 10t … (k) dt2 dt Nonhomogeneous differential equation i(t) = iC + iP Complementary solution Characteristic equation is s2i + 3si + 2i = 0 s2 + 3s + 2 = 0 A B C -b ± √b2 – 4ac s1, s2 = 2a Putting corresponding values we get s1, s2 = s1, s2 = s1, s2 = s1, s2 =

94

1 3 2

-3 ± √32 – 4(1)(2) 2(1) -3 ± √9 – 8 2(1) -3 ± √1 2(1) -3 ± 1

2 s1, s2 = (-3 + 1)/2, (-3 – 1)/2 s1, s2 = (-1), (-2) General solution: v(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e(-1)t + K2e(-2)t … (i) Particular solution From table 6.2 iP = A sin 10t + B cos 10t Putting the value of iP in (k) d2 d [A sin 10t + B cos 10t] + 3 [A sin 10t + B cos 10t] + 2[A sin 10t + B cos 10t] = dt2 dt 10 sin 10t -100A sin 10t – 100B cos 10t + 3[10A cos 10t - 10B sin 10t] + 2A sin 10t + 2B cos 10t = 10 sin 10t

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

95

-100A sin 10t – 100B cos 10t + 30A cos 10t - 30B sin 10t + 2A sin 10t + 2B cos 10t = 10 sin 10t Equating coefficients Sin -100A – 30B + 2A = 10 -98A – 30B = 10 Cos -100B + 30A + 2B = 0 -98B + 30A = 0 B = 0.306A -98A – 30(0.306A) = 10 -98A – 9.18A = 10 -107.18A = 10 A = -0.094 B = 0.306(-0.094) B = -0.029 Now iP = -0.094 sin 10t + (-0.029) cos 10t iP = -0.094 sin 10t - 0.029 cos 10t i(t) = K1e(-1)t + K2e(-2)t + [-0.094 sin 10t - 0.029 cos 10t] i(t) = K1e(-1)t + K2e(-2)t - 0.094 sin 10t - 0.029 cos 10t At t = 0+ i(0+) = K1e(-1)0+ + K2e(-2)0+ - 0.094 sin 10(0+) - 0.029 cos 10(0+) … (c) 2 = K1e0 + K2e0 - 0.094 sin 0 - 0.029 cos 0 2 = K1(1) + K2(1) - 0.029 (1) 2 = K1 + K2 - 0.029 2.029 = K1 + K2 … (y) Differentiating eq. (c) with respect to ‘t’ di (0+) = -K1e-(0+) - 2K2e(-2)0+ - 0.094(10) cos 10(0+) + 0.029(10) sin 10(0+) dt -1 = -K1e0 - 2K2e0 - 0.94 cos 0 + 0.029(10) sin 0 -1 = -K1(1) - 2K2(1) - 0.94 (1) -1 = -K1 - 2K2 - 0.94 0.06 = K1 + 2K2 From eq. (y) K1 = 2.029 – K2 0.06 = [2.029 – K2] + 2K2 0.06 = 2.029 – K2 + 2K2 -1.969 = K2 Putting the value of K2 in (y) 1.029 = K1 + (-1.969) 1.029 + 1.969 = K1 2.998 = K1 i(t) = 2.998e(-1)t + (-1.969)e(-2)t - 0.094 sin 10t - 0.029 cos 10t

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

96

i(t) = 2.998e(-1)t - 1.969e(-2)t - 0.094 sin 10t - 0.029 cos 10t (d) d2q

dq +5 + 6q = te-t … (m) dt2 dt Nonhomogeneous differential equation q(t) = qC + qP Complementary solution Characteristic equation is s2q + 5sq + 6q = 0 s2 + 5s + 6 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get

s1, s2 = s1, s2 = s1, s2 =

-5 ± √52 – 4(1)(6) 2(1) -5 ± √25 – 24 2(1) -5 ± √1 2(1)

s1, s2 =

-5 ± 1

2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: q(t) = K1es1t + K2es2t Putting corresponding values we get q(t) = K1e-2t + K2e-3t

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Particular solution qP = (At + B)te-t ⇒ multiplied by ‘t’ because e-t is repeating in both iP & iC Putting the value of iP in eq. m d2 d [(At + B)te-t] + 5 [(At + B)te-t] + 6[(At + B)te-t] = te-t dt2 dt At2e-t - 2Ate-t – 2Ate-t + 2Ae-t + Bte-t – Be-t – Be-t + 5[-At2e-t + 2Ate-t - Bte-t + Be-t] + 6[At2e-t + Bte-t] = te-t At2e-t - 2Ate-t – 2Ate-t + 2Ae-t - Bte-t – Be-t – Be-t - 5At2e-t + 10Ate-t - 5Bte-t + 5Be-t + 6At2e-t + 6Bte-t = te-t Equating coefficients t2e-t: A – 5A + 6A = 0 2A = 0 A=0 te-t: -2A – 2A + 10A = 1 6B = 3 B = 0.5 e-t: 2 – B – B + 5B = 0 2 + 3B = 0 B = -0.667 When B = 0.5 qP = (0t + (0.5))te-t qP = 0.5te-t When B = -0.667 qP = (0t + (-0.667))te-t qP = -0.667te-t q(t) = K1e-2t + K2e-3t - 0.667te-t, K1e-2t + K2e-3t + 0.5te-t At t = 0+ q(0+) = K1e-2(0+) + K2e-3(0+) - 0.667(0+)e-(0+), K1e-2(0+) + K2e-3(0+) + 0.5(0+)e-(0+) 2 = K1e0 + K2e0, K1e0 + K2e0 2 = K1(1) + K2(1), K1(1) + K2(1) 2 = K1 + K2, K1 + K2 … (y) q(t) = K1e-2t + K2e-3t - 0.667te-t Differentiating with respect to ‘t’ at t = 0+ dq (0+) = -2K1e-2(0+) -3K2e-3(0+) - 0.667[-(0+)e-(0+) + e-(0+)] dt -1 = -2K1e0 -3K2e0 - 0.667[e0] -1 = -2K1(1) -3K2(1) - 0.667 -1 = -2K1 -3K2 - 0.667 0.333 = 2K1 + 3K2 From eq. (y) K1 = 1 – K2 0.333 = 2[2 – K2] + 3K2

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0.333 = 4 – 2K2 + 3K2 0.333 = 4 + K2 K2 = 0.333 - 4 K2 = -3.667 1 = K1 + (-3.667) K1 = 4.667 q(t) = 4.667e-2t + (-3.667)e-3t - 0.667te-t q(t) = 4.667e-2t - 3.667e-3t - 0.667te-t (e) d2V

dV

+5 + 6V = e-2t + 5e-3t … (n) dt dt Nonhomogeneous differential equation V(t) = VC + VP Complementary solution Characteristic equation is s2V + 5sV + 6V = 0 s2 + 5s + 6 = 0 a b c 2

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -5 ± √52 – 4(1)(6) s1, s2 = 2(1) -5 ± √25 – 24 s1, s2 = 2(1) -5 ± √1 s1, s2 = 2(1) -5 ± 1 s1, s2 = 2 s1, s2 = (-5 + 1)/2, (-5 - 1)/2 s1, s2 = (-4)/2, (-6)/2 s1, s2 = -2, -3 General solution: V(t) = K1es1t + K2es2t

1 5 6

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Putting corresponding values we get V(t) = K1e-2t + K2e-3t Particular solution VP = Ce-2t + Ce-3t Substituting the value of VP in eq. (n) d2 d -2t -3t [Ce + Ce ] + 5 [Ce-2t + Ce-3t] + 6[Ce-2t + Ce-3t] = e-2t + 5e-3t dt2 dt -2t -3t 4Ce + 9Ce + 5[-2Ce-2t – 3Ce-3t] + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t 4Ce-2t + 9Ce-3t - 10Ce-2t – 15Ce-3t + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t Equating coefficients e-2t 4C – 10C + 6C = 1 4C – 10C + 6C = 1 C=0 VP = (0)e-2t + (0)e-3t VP = 0 V(t) = K1e-2t + K2e-3t + 0 V(t) = K1e-2t + K2e-3t … (k) At t = 0+ V(0+) = K1e-2(0+) + K2e-3(0+) 2 = K1e0 + K2e0 2 = K1(1) + K2(1) 2 = K1 + K2 … (A) Differentiating eq. (k) with respect to ‘t’ dV (0+) = -2K1e-2(0+) - 3K2e-3(0+) dt -1 = -2K1e0 - 3K2e0 -1 = -2K1(1) - 3K2(1) -1 = -2K1 - 3K2 … (B) K1 = 2 – K2 -1 = -2[2 – K2] - 3K2 … (B) -1 = -4 + 2K2 - 3K2 -3 = K2 1 = K1 + (-3) … (A) 4 = K1 V(t) = 4e-2t + (-3)e-3t … (k) V(t) = 4e-2t - 3e-3t Q#6.21: Solve the differential equation d3i d2i di 2 +9 + 13 + 6i = K0te-tsint … (u) dt dt dt 3 2 2s i + 9s i + 13si + 6i = 0

99

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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X-TICS EQUATION: 2s3 + 9s2 + 13s + 6 = 0 We can find roots using synthetic division 2 -1 2

9

13

6

-2

-07

-6

6

0

7

s = -1 So s + 1 is a factor. (s + 1)(2s2 + 7s + 6) = 0 2s2 + 7s + 6 = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -7 ± √72 – 4(1)(6) s1, s2 = 2(1) -7 ± √49 – 24 s1, s2 = 2(1) -7 ± √25 s1, s2 = 2(1) -7 ± 5 s1, s2 = 2 s1, s2 = (-7 + 5)/2, (-7 – 5)/2 s1, s2 = (-2)/2, (-12)/2 s1, s2 = -1, -6 General solution: i(t) = K1es1t + K2es2t Putting corresponding values we get i(t) = K1e-1t + K2e-6t So –1, -1, -6 are roots of the X-TICS equation. General solution: i(t) = K1e-1t + K2e-1t + K3e-6t … (i)

Remainder is zero.

2 7 6

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Particular olution iP = Fte-tcos t + Gte-tsin t Putting the value of iP in (u) Here in eq. (u) i = iP Simplification Ist derivative of iP: = F(1)e-tcos t + Ft(-e-t)cos t + Fte-t(-sin t) + G(1)e-tsin t + Gt(-e-t)sin t + Gte-tcos t = Fe-tcos t - Fte-tcos t - Fte-tsin t + Ge-tsin t - Gte-tsin t + Gte-tcos t 2nd derivative of iP: = F[-e-tsin t – e-tcos t] – F[e-tcos t - te-tcos t - te-tsin t] – F[e-tsin t - te-tsin t + te-tcos t] + G[e-tcos t – e-tsin t] – G[e-tsin t - te-tsin t + te-tcos t] + G[e-tcos t - te-tcos t - te-tsin t] = -2Fe-tsin t – 2Fe-tcos t + 2Fte-tsin t + 2Ge-tcos t – 2Ge-tsin t - 2Gte-tcos t 3rd derivative of iP: = -2F[e-tcos t – e-tsin t] - 2F[-e-tsin t – e-tcos t] + 2F[e-tsin t - te-tsin t + te-tcos t] + 2G[-e-tsin t – e-tcos t] – 2G[e-tcos t – e-tsin t] – 2G[e-tcos t - te-tcos t - te-tsin t] = 6Fe-tsin t – 2Fte-tsin t + 2Fte-tcos t – 6Ge-tcos t + 2Gte-tcos t + 2Gte-tsin t 2[6Fe-tsin t – 2Fte-tsin t + 2Fte-tcos t – 6Ge-tcos t + 2Gte-tcos t + 2Gte-tsin t] + 9[-2Fe-tsin t – 2Fe-tcos t + 2Fte-tsin t + 2Ge-tcos t – 2Ge-tsin t - 2Gte-tcos t] + 13[Fe-tcos t - Fte-tcos t - Fte-tsin t + Ge-tsin t - Gte-tsin t + Gte-tcos t] + 6[Fte-tcos t + Gte-tsin t] = K0te-tsin t 12Fe-tsin t – 4Fte-tsin t + 4Fte-tcos t – 12Ge-tcos t + 4Gte-tcos t + 4Gte-tsin t - 18Fe-tsin t – 4Fe-tcos t + 4Fte-tsin t + 4Ge-tcos t – 4Ge-tsin t - 4Gte-tcos t + 13Fe-tcos t - 13Fte-tcos t - 13Fte-tsin t + 13Ge-tsin t - 13Gte-tsin t + 13Gte-tcos t + 6Fte-tcos t + 6Gte-tsin t = K0te-tsin t Comparing coefficients e-tsin t: 12F – 18F – 4G + 13G = 0 -6F + 9G = 0 -2F + 3G = 0 … (7) te-tsin t: -4F + 4G + 4F – 13F – 13G + 6G = K0 -3G – 13F = K0 … (8) From eq. (7) F = 1.5G put in (8) -3G – 13(1.5G) = K0

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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-3G – 19.5G = K0 -22.5G = K0 G = -0.045K0 Put in (7) -2F + 3(-0.045K0) = 0 -2F – 0.135K0 = 0 F = -0.068K0 iP = (-0.068K0)te-tcos t + (-0.045K0)te-tsin t iP = -0.068K0te-tcos t - 0.045K0te-tsin t i(t) = K1e-1t + K2e-1t + K3e-6t + (-0.068K0te-tcos t - 0.045K0te-tsin t) i(t) = K1e-1t + K2e-1t + K3e-6t - 0.068K0te-tcos t - 0.045K0te-tsin t … (9) At t = 0+ i(0+) = K1e-1(0+) + K2e-1(0+) + K3e-6(0+) - 0.068K0(0+)e-(0+)cos (0+) - 0.045K0(0+)e-(0+)sin (0+) 1 = K1e0 + K2e0 + K3e0 1 = K1(1) + K2(1) + K3(1) 1 = K1 + K2 + K3 Differentiating eq. (9) with respect to ‘t’ di (0+) = -K1e-(0+) - K2e-1(0+) - 6K3e-6(0+) - 0.068K0[e-(0+)cos (0+) dt – (0+)e-(0+)cos (0+) – (0+)e-(0+)sin (0+)] - 0.045K0[e-(0+)sin (0+) – (0+)e-(0+)sin (0+) + (0+)e(0+) cos (0+)] After simplification -1 = -K1 – K2 – 6K3 - 0.068K0 -1 = -K1 – K2 – 6K3 - 0.068K0 1 = K1 + K2 + 6K3 + 0.068K0 Q#6.22: A special generator has a voltage variation given by the equation V(t) = t V, where t is the time in seconds and t ≥ 0. This generator is connected to an RL series circuit, where R = 2 Ω and L = 1 H, at time t = 0 by the closing of a switch. Find the equation for the current as a function of time i(t).

K

L + -

V(t)

i R

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

At t ≥ 0 According to KVL di L + iR = V(t) … (c) dt Putting corresponding values di + 2i = t dt Nonhomogeneous differential equation V(t) = VC + VP Complementary solution Characteristic equation is si + 2i = 0 s+2=0 s = -2 iC = Kest iC = Ke(-2)t iC = Ke-2t Particular solution From table 6.2 A1tn B0tn + B1tn- 1 + … + Bn – 1t + Bn iP = At + B … (ii) Putting the value of iP from (ii) into eq. (c) d [At + B] + 2[At + B] = t dt After simplifications A + 2At + 2B = t Equating coefficients t: 2A = 1 A = 0.5 Constants: A + 2B = 0 0.5 + 2B = 0 B = -0.25 iP = 0.5t + (-0.25) … (ii) iP = 0.5t - 0.25 i(t) = Ke-2t + 0.5t - 0.25 At t = 0+

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Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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i(0+) = Ke-2(0+) + 0.5(0+) - 0.25 0 = Ke0 - 0.25 ⇒ i(0+) = i(0-) = 0 (due to inductor) 0 = K(1) - 0.25 K = 0.25 i(t) = 0.25e-2t + 0.5t - 0.25 Q#6.23: A bolt (flash of lighting) of lightning having a waveform which is approximated as V(t) = te-t strikes a transmission line having resistance R = 0.1 Ω and inductance L = 0.1 H. Find i(t) =? Solution:

+ -

At t ≥ 0 According to KVL di L + iR = V(t) dt Putting corresponding values di (0.1) + (0.1)i = te-t … (h) dt Nonhomogeneous differential equation V(t) = VC + VP Complementary solution Characteristic equation is 0.1si + 0.1i = 0 s+1=0 s = -1 iC = Kest iC = Ke(-1)t iC = Ke-1t Particular solution From table 6.2 Particular solution iP = (At + B)te-t ⇒ multiplied by ‘t’ because e-t is repeating in both iP & iC Putting the value of qP in eq. (h) 0.1[-At2e-t + 2Ate-t - Bte-t + Be-t] + 0.1[At2e-t + Bte-t] = te-t -0.1At2e-t + 0.2Ate-t – 0.1Bte-t + 0.1Be-t + 0.1At2e-t + 0.1Bte-t] = te-t

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} Equating coefficients: t2e-t: -0.1A + 0.1A = 0 te-t: 0.2A – 0.1B + 0.1B = 1 0.2A = 1 A=5 e-t: 0.1B = 0 B=0 iP = (5t + 0)te-t iP = 5t2e-t i(t) = Ke-1t + 5t2e-t i(t) = Ke-t + 5t2e-t At t = 0+ i(0+) = Ke-(0+) + 5(0+)2e-(0+) 0 = Ke0 0=K i(t) = 0e-1t + 5t2e-t i(t) = 5t2e-t i(0+) = i(0-) = 0 (due to inductor) Q#6.24: In the network of the figure, the switch K is closed at t = 0 with the capacitor initially unenergized. For the numerical values given, find i(t).

+ -

Here VC(0-) = VC(0+) = 0 V & 10e-(0+) sin (0+) u(0+) i(0+) = 10 10e0 sin (0) u(0+) i(0+) = 10 i(0+) = 0 Amp. For t ≥ 0, KVL

105

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

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1 10i + -6

(10(10 )) 1 10i +

∫ idt = 10e-t sin t u(t) ∫ idt = 10e-t sin t

(10(10-6)) After simplification i + 104∫ idt = e-t sin t Differentiating with respect to ‘t’ di + 104i = e-t (cost - sin t) … (1) dt It is a nonhomogeneous differential equation Characteristic equation is si + 104i = 0 s + 104 = 0 s = -104 Therefore complementary function is iC = Ke-10, 000t The form of particular integral is From table 6.2 iP = (Acost + Bsint)e-t Putting the value of iP in eq. (1) After simplification -Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000[(Acost + Bsint)e-t] = e-t (cost - sin t) -Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000Acoste-t + 10, 000Bsinte-t = e-t cost - e-t sin t Equating coefficients e-tsint: -A – B + 10, 000B = -1 -A + 9999B = -1 … (2) e-tcost: -A + B + 10, 000A = 1 B + 9999A = 1 … (3) From (2) A = 1 + 9999B B + 9999[1 + 9999B] = 1 … (3) B + 9999 + 9999(9999)B = 1 B + 99980001B = -9998 B[1 + 99980001] = -9998 B = -9998 B = -10-4 A = 1 + 9999(-1/10000) A = 1 – 0.9999 A = 10-4 iP = (10-4cost + (-10-4)sint)e-t

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

107

iP = (10-4cost - 10-4sint)e-t Complete solution is i(t) = Ke-10, 000t + (10-4cost - 10-4sint)e-t At t = 0+ i(0+) = Ke-10, 000(0+) + (10-4cos(0+) - 10-4sin(0+))e-(0+) 0 = Ke0 + (10-4(1) - 10-4(0))e0 0 = K(1) + (10-4(1) - 10-4(0))(1) 0 = K + 10-4 K = -10-4 i(t) = (-10-4)e-10, 000t + (10-4cost - 10-4sint)e-t Q#6.25: In the network shown in the accompanying figure, a steady state is reached with the switch K open. At t = 0, the switch is closed. For the element values given, determine the current, i(t) for t ≥ 0. Solution:

+ -

100 sin 377 t

R = 1000 Ω

Equivalent circuit before switching: By phasors: V = 100∠ 00 jXL = jω L jXL = j(377)(1) jXL = j377 1 -jXC = -j ωC -jXC = -j5305.04 Therefore V I= R + jXL - jXC After simplification I = 0.0198∠ 78.530 That is i(t) = 0.0198sin(377t + 78.530) Also VC = I(-jXC) VC = (0.0198∠ 78.530)(5305.04∠ -900) VC = 105.040∠ -11.470

0.5 µ F L=1H

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

108

Therefore VC(t) = 105.040sin(377t + (-11.470)) VC(t) = 105.040sin(377t - 11.470) At t = 0+ i(0+) = 0.0198sin(377(0+) + 78.530) i(0+) = 0.0198sin(78.530) i(0+) = 0.0198[0.981] i(0+) = 0.019 amp At t = 0+ VC(0+) = 105.040sin(377(0+) - 11.470) VC(0+) = 105.040sin(-11.470) VC(0+) = 105.040[-sin(11.470)] VC(0+) = 105.040[-0.199] VC(0+) = -20.903 Volts For t ≥ 0 According to KVL di 1 L + ∫ idt = 100sin377t … (d) dt C Differentiating with respect to ‘t’ d2i i L + = 100(377)cos377t 2 dt C d2i i 37700cos377t + = dt2 LC L 2 di i 37700cos377t + = dt2 (1)(0.5(10-6)) 1 d2i + 2(106)i = 37700cos377t … (5) dt2 s2i + 2(106)i = 37700cos377t Complementary solution: s2 + 2(106) = 0 s = ± j1414.21 iC = K1cos1414.21 t + K2sin1414.21 t Particular solution: iP = Acos377t + Bsin377t Putting the value of iP in eq. (5) After simplification -(377)2Acos377t – (377)2Bsin(377)t + 2(106)[Acos(377)t + Bsin(377)t] = 37700cos377t -(377)2Acos377t – (377)2Bsin377t + 2000000Acos377t + 2000000Bsin377t = 37700cos377t Equating coefficients:

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} cos377t: -(377)2A + 2000000A = 37700 -142129A + 2000000A = 37700 1857871A = 37700 A = 0.020 sin377t: –(377)2B + 2000000B = 0 B=0 iP = 0.02cos377t + (0)sin377t iP = 0.02cos377t Complete solution: i(t) = K1cos1414.21 t + K2sin1414.21 t + 0.02cos377t … (Q) At t = 0+ i(0+) = K1cos1414.21(0+) + K2sin1414.21(0+) + 0.02cos377(0+) 0.019 = K1cos(0) + 0.02cos(0) = K1(1) + 0.02(1) = K1 + 0.02 K1 = 0.019 – 0.02 K1 = -0.001 Differentiating eq. (Q) with respect to ‘t’ di(t) = -1414.21K1sin1414.21 t + 1414.21K2cos1414.21 t + 0.02(-377s)sin377t dt From differential eq. (d) at t = 0+ di(0+) + VC(0+) = 100sin377t … (d) dt di(0+) + VC(0+) = 100sin377(0+) dt di(0+) + (-20.903) = 0 dt di(0+) = 20.903 dt 20.903 = -1414.21K1sin1414.21(0+) + 1414.21K2cos1414.21(0+) + 0.02(-377s)sin377(0+) 20.903 = -1414.21K1sin0 + 1414.21K2cos0 + 0.02(-377s)sin0 20.903 = 1414.21K2(1) + 0 20.903 = 1414.21K2 K2 = 0.015 Therefore i(t) = -0.001cos1414.21 t + 0.015sin1414.21 t + 0.02cos377t

109

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

110

Q#6.26: In the network shown in figure, a steady state is reached with the switch K open. At t = 0, the value of the resistor R is changed to the critical value, Rcr defined by equation (6.88). For the element values given, determine the current i(t) for t ≥ 0. Solution: L Rcr = 2 C Rcr = 2(1414.214) Rcr = 2828.428 Ω For t ≥ 0 According to KVL di 1 Rcri + L + ∫ idt = 100sin377t … (x) dt C Differentiating with respect to ‘t’ di d2i i Rcr + L + = 37700cos377t 2 dt dt C After simplification di d2i 2828.428 + (1) + 2(106)i = 37700cos377t … (w) 2 dt dt Characteristic equation is s2i + 2828.428si + 2(106)i = 0 s2 + 2828.428s + 2(106) = 0 A B C

s1, s2 =

-b ± √b2 – 4ac

2a Putting corresponding values we get -2828.428 ± √2828.4282 – 4(1)(2000000) s1, s2 = 2(1) -2828.428 ± √8000004.952 – 8000000 s1, s2 = 2(1) -2828.428 ± √4.952 s1, s2 = 2(1)

1 2828.428 2(106)

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

s1, s2 =

111

-2828.428 ± 2.225

2(1) s1, s2 = (-2828.428 + 2.225)/2, (-2828.428 - 2.225)/2 s1, s2 = -1413.102, -1415.326 i(t) = K1e-1413.102t + K2e-1415.326t Form of particular integral is iP = Acos377t + Bsin377t Putting the value of iP in eq. (w) -(377)2Acos377t – (377)2Bsin377t + 2828.428[-377Asin377t + B377sin377t] + 2000000[Acos377t + Bsin377t] = 37700cos377t -142129Acos377t – 142129Bsin377t - 1066317.356Asin377t + 1066317.356Bsin377t + 2000000Acos377t + 2000000Bsin377t = 37700cos377t Equating coefficients: cos377t: -142129A + 2000000A = 37700 A = 0.020 sin377t: –142129B - 1066317.356A + 1066317.356B + 2000000B = 0 2924188.356B - 1066317.356(0.020) = 0 2924188.356B = 1066317.356(0.020) 2924188.356B = 21326.34712 B = 0.007 iP = 0.020cos377t + 0.007sin377t Complete solution: i(t) = K1e-1413.102t + K2e-1415.326t + 0.020cos377t + 0.007sin377t … (a) At t = 0+ i(0+) = K1e-1413.102(0+) + K2e-1415.326(0+) + 0.020cos377(0+) + 0.007sin377(0+) 0.019 = K1e0 + K2e0 + 0.020 0.019 = K1(1) + K2(1) + 0.020 -0.001 = K1 + K2 … (z) Differentiating eq. (a) with respect to ‘t’ at t = 0+ di (0+) = -1413.102K1e-1413.102(0+) - 1415.326K2e-1415.326(0+) + 0.020(-377)sin377(0+) + dt 0.007(377)cos377(0+) … (g) At t = 0+ eq. (x) gives di (0+) = -32.84 dt -32.84 = -1413.102K1e-1413.102(0+) - 1415.326K2e-1415.326(0+) + 0.020(-377)sin377(0+) + 0.007(377)cos377(0+) … (g)

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-32.84 = -1413.102K1e0 - 1415.326K2e0 + 0.007(377)cos 0 -32.84 = -1413.102K1(1) - 1415.326K2(1) + 0.007(377)(1) -32.84 = -1413.102K1 - 1415.326K2 + 2.639 -32.84 – 2.639 = -1413.102K1 - 1415.326K2 -35.479 = -1413.102K1 - 1415.326K2 35.479 = 1413.102K1 + 1415.326K2 From eq. (z) K1 = -0.001 – K2 35.479 = 1413.102[-0.001 – K2] + 1415.326K2 35.479 = 1413.102[-0.001] – 1413.102K2 + 1415.326K2 35.479 = -1.413102 + 2.224K2 K2 = 16.588 K1 = -0.001 – 16.588 K1 = -16.589 i(t) = [-16.589]e-1413.102t + [16.588]e-1415.326t + 0.020cos377t + 0.007sin377t Q#6.27: Consider the network shown in figure P6.24. The capacitor has an initial voltage, VC = 10 V. At t = 0, the switch K is closed. Determine i(t) for t ≥ 0. Solution: Here VC(0-) = VC(0+) = 10 V V(t) – VC(t) i(t) = 10 V(0+) – VC(0+) i(0+) = 10 V(0+) = 10e-(0+) sin (0+) u(0+) V(0+) = 10(1)(0)(1) V(0+) = 0 Volts 0 – 10 i(0+) = 10 i(0+) = -1 amp. For t ≥ 0, KVL 1 10i + ∫ idt = 10e-t sin t u(t) -6 (10(10 )) 1 10i + ∫ idt = 10e-t sin t (10(10-6)) After simplification i + 104∫ idt = e-t sin t Differentiating with respect to ‘t’ di

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} + 104i =

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e-t (cost - sin t) … (1)

dt It is a nonhomogeneous differential equation Characteristic equation is si + 104i = 0 s + 104 = 0 s = -104 Therefore complementary function is iC = Ke-10, 000t The form of particular integral is From table 6.2 iP = (Acost + Bsint)e-t Putting the value of iP in eq. (1) After simplification -Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000[(Acost + Bsint)e-t] = e-t (cost - sin t) -Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000Acoste-t + 10, 000Bsinte-t = e-t cost - e-t sin t Equating coefficients e-tsint: -A – B + 10, 000B = -1 -A + 9999B = -1 … (2) e-tcost: -A + B + 10, 000A = 1 B + 9999A = 1 … (3) From (2) A = 1 + 9999B B + 9999[1 + 9999B] = 1 … (3) B + 9999 + 9999(9999)B = 1 B + 99980001B = -9998 B[1 + 99980001] = -9998 B = -9998 B = -10-4 A = 1 + 9999(-1/10000) A = 1 – 0.9999 A = 10-4 iP = (10-4cost + (-10-4)sint)e-t iP = (10-4cost - 10-4sint)e-t Complete solution is i(t) = Ke-10, 000t + (10-4cost - 10-4sint)e-t At t = 0+ i(0+) = Ke-10, 000(0+) + (10-4cos(0+) - 10-4sin(0+))e-(0+) -1 = Ke0 + (10-4(1) - 10-4(0))e0 -1 = K(1) + (10-4(1) - 10-4(0))(1) -1 = K + 10-4 K = -1 - 10-4 i(t) = (-1 - 10-4)e-10, 000t + (10-4cost - 10-4sint)e-t

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Q#6.28: The network of the figure is operating in the steady state with the switch K open. At t = 0, the switch is closed. Find an expression for the voltage, V(t) for t ≥ 0. Solution:

Equivalent circuit before switching:

V(0-) = V(0+) = I0sin ω t [R1 + R2]

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)}

Equivalent circuit after switching:

For t ≥ 0 According to KCL dV(t) V(t) C + = I0sin ω t dt R1 dV(t) V(t) I0sin ω t + = … (u) dt CR1 C It is a nonhomogeneous differential equation Characteristic equation is 1 sV(t) + V(t) = 0 CR1 1 s+ =0 CR1 1 s=CR1 Therefore complementary function is VC = Ke-(1/CR1)t The form of particular integral is VP = Asin ω t + Bcos ω t Putting the value of VP in eq. (u) ω Acos ω t - ω Bsin ω t + [1/CR1][Asin ω t + Bcos ω t] = (I0sin ω t)/C

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ω Acos ω t - ω Bsin ω t + [1/CR1]Asin ω t + [1/CR1]Bcos ω t = (I0sin ω t)/C Equating coefficients: cos ω t: ω A + [1/CR1]B = 0 ⇒ A = -B/ω CR1 sin ω t: -ω B + [1/CR1]A = I0/C -ω B + [1/CR1][-B/ω CR1] = I0/C B = I0/C[-ω - 1/C2R12] A = -[I0/C[-ω - 1/C2R12]]/ω CR1 Complete solution V(t) = Ke-(1/CR1)t + [-[I0/C[-ω - 1/C2R12]]/ω CR1]sin ω t + [I0/C[-ω - 1/C2R12]]cos ω t At t = 0+ V(0+) = Ke-(1/CR1)0+ + [-[I0/C[-ω - 1/C2R12]]/ω CR1]sin ω (0+) + [I0/C[-ω - 1/C2R12]]cos ω (0+) I0sin ω t [R1 + R2] = K(1) + [I0/C[-ω - 1/C2R12]] I0sin ω t [R1 + R2] = K + [I0/C[-ω - 1/C2R12]] K = I0sin ω t [R1 + R2] - [I0/C[-ω - 1/C2R12]] V(t) = [I0sin ω t [R1 + R2] - [I0/C[-ω - 1/C2R12]]]e-(1/CR1)t + [-[I0/C[-ω - 1/C2R12]]/ω CR1]sin ω t + [I0/C[-ω - 1/C2R12]]cos ω t Q#6.29: Consider a series RLC network which is excited by a voltage source. Determine the characteristic equation. Locus of the roots of the equation. Plot the roots of the equation. Solution: