# Differential Equations - Solved Assignments - Semester Spring 2005

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Differential Equations - Solved Assignments - Semester Spring 2005...

#### Description

Assignment 1 (Spring 2005) Maximum Marks 60 Due Date 07, April 2005 Assignment Weight age 2%

Question 1

(a) Define initial value and boundary value problem elaborate with the help of examples (at least one). 05 Solution INITIAL VALUE PROBLEM Differential equation of first order or greater in which the dependent variable y or its derivative are specified at one points such as dy  f  x, y  dx y  x0   y0 If equation of the second order d2y dy a2  x  2  a1  x   a0  x  y  g  x  dx dx ' y  x0   y0 , y  x0   y '1 Where y0 and y '1 are arbitrary constants

Is called the initial value problem and the y  x0   y0 , y '  x0   y '1 is called the initial conditions BOUNDARY VALUE PROBLEM Differential equation of order two or greater in which the dependent variable y or its derivative are specified at different points such as d2y dy  a1  x   a0  x  y  g  x  2 dx dx y  x0   y0 , y  x1   y1

a2  x 

Where y0 and y '1 are arbitrary constants

Is called the boundary value problem and the y  x0   y0 , y  x1   y1 is called the boundary conditions

NOT THE PART OF THE SOLUTION (MOTIVATION OF THE CONDITIONS)

Basically when we solve the differential equations then we find some constant in the solution so we need to elaborate these constants to solve this problem we formulate two kinds of conditions(given above) which basically helps us to elaborate the constants finding in the solution of the differential equation. Still the question is how many constants are present in the differential equation. Is there any hard and fast rule? yes! It depends that what is the order of the differential equation. If we have the first order DE then its solution contains must one constant similarly if second order DE then its solution contains must two constants……and so on and definitely to elaborate the constant we always need same number of condition.

Question 1

(b) Solve the following differential equation. xdx  ydy 

Solution

xdy  ydx  0; y (0)  1 x2  y 2

10

xdx  ydy  (

xdy  ydx )  0 , y 0  1 x2  y 2

  y  x  dx   y  2 x 2  dy  0 2  x y  x  y2    f y  M ( x, y )  x  2 x x  y2 f x  N ( x, y )=y  2 y x  y2  y  M x 2  x  y2    x 3  xy 2  y  M   2 2  x y 

 x  ,N  y 2  x  y2    x2 y  y3  x  N   2 2  x y 

2 2 3 2 M  2 xy  1  x  y   2 y  x  xy  y   2 y  x2  y 2  3 3 2 2 3 3 2 M  2 x y  2 xy  x  y    2 x y  2 xy  2 y   2 y  x2  y 2 

M 2 x 3 y  2 xy 3  x 2  y 2  2 x 3 y  2 xy 3  2 y 2  2 y  x2  y 2  M y2  x2  now y  x 2  y 2 2 2 2 2 3 N  2 xy  1  x  y   2 x  x y  y  x   2 x  x2  y 2  3 3 2 2 3 3 2 N  2 x y  2 xy  x  y    2 x y  2 xy  2 x   2 x  x2  y 2 

N 2 x 3 y  2 xy 3  x 2  y 2  2 x 3 y  2 xy 3  2 x 2  2 x  x2  y 2  N y2  x2 M   2 2 2 x  x  y  y

so our given equation is exact now as we know f y  M ( x, y )  x  2 x x  y2 Integrating both sides with respect to “x”

f ( x, y )   Mdx  y  1 f ( x, y )    x  2 dx   xdx  y  2 dx   ( y ) 2  x y  x  y2  1 x x2 f ( x, y )   y   tan 1     ( y ) 2  y  y dx 1 x sin ce  2  tan 1   2 x a a a f ( x, y )  f  y

x x2  tan 1     ( y )              (1) 2  y

1  x  '     ( y) x2  y 2  1 2 y

f y2  x  '  2     ( y) y y  x2  y 2  f x  2   ' ( y) 2 y y  x as we knowthat f x  N ( x, y )  y  2 y x  y2 we will get x x  2   ' ( y) 2 x y y  x2  '( y )  y y

2

y2 2 now putting this value in equation no1

 ( y) 

so f ( x, y ) 

x2  y  y2  tan 1    2 x 2

x2  y  y2  tan 1    2 x 2 applying initial condition we get

c

1 2 2  1 x2  y  y2    tan 1    2 2 2 x 2

c

Question 2

(a) Define the order and degree of DE also write down the order and degree of following DE.

1 x 2 y"  xy '  ( x 2 1) y  0  2 19 x 2  y"'   4 xy '  (4 x 2  25)  0 2

04

Solution Order of Differential Equation: Order of the differential equation is the highest order derivative in a differential equation e.g. y''+(y')3 =0 has order two. Degree of the Differential Equation: Degree of the differential equation is the power of the highest order derivative in the differential e.g. y''+y=2 has degree one. Although, (y'') 3+y'+y=0 has degree three. Order (1) 2 (2) 3 Degree (1) 1 (2) 2

Question 2

(b) Find I.F from 3rd and 4th method of exactness i.e. (xM+yN) & (xM-yN) respectively if possible. If not then explain the reason.  x3 y  4xy3  dx   x5  6x4 y  dy  0 Solution Integrating factor for 3rd case can not be found due to that xM  yN  0 is not homogenous So first case is not applicable as M= x3 y  4 xy 3 , N= x5  6 x 4 y xM  yN  x  x3 y  4 xy 3  +y  x5  6 x 4 y 

xM  yN  x 4 y  4 x 2 y 3 +x5 y  6 x 4 y 2 To check it is homogeneous

05

Say xM  yN  x 4 y  4 x 2 y 3 +x 5 y  6 x 4 y 2  f  x, y  f  tx, ty   t n f  x, y  , for some real number n but f  tx, ty   t 5 x 4 y  4t 5 x 2 y 3 +t 6 x 5 y  6t 6 x 4 y 2 f  tx, ty   t 5  x 4 y  4 x 2 y 3 +tx 5 y  6tx 4 y 2  f  tx, ty   t 5 f  x, y 

For 4th case it is not possible because condition yf ( xy)dx  xg ( xy)dy  0 is not being fulfilled here. As  x3 y  4 xy3  dx   x5  6 x4 y  dy  0 y  x3  4 xy 2  dx  x  x 4  6 x3 y  dy  0

But x3  4 xy 2 and x 4  6 x3 y are not the function of xy NOTE (RMEMBER)

f  xy  Means each term should have the product of xy e.g. f  xy   xy  x 2 y 2 And f  xy   xy  x2 y 2  x4 y 4 Both are functions of xy Question 2

(c) Define ordinary and partial differential equation as well as check whether the following are ordinary differential equation or partial differential equation. d 2 x dy dz    2 Where x, y and z are dependent variable and u is independent variable du 2 du du d 2x dx Where x is dependent variable and u is independent variable  2 2  3  4 du du 2 x 2 x Where x is dependent variable, u and v are independent variable 3   2  2 3 u v  2u  2 v Where u, v are dependent variable, x and y are independent variable  4 2  2  4 x y

1 4

Solution ORDINARY DIFFERENTIAL EQUATION (ODE)

06

If an equation contains only ordinary derivatives of one or more dependent variables depending on only one independent variable is known as ODE. PARTIAL DIFFERENTIAL EQUATION (PDE) If an equation contains partial derivatives of one or more dependent variables depending on two or more independent variables is known as PDE (1) ODE (2) ODE (3) PDE (4) PDE Question 3

(a) Define explicit and implicit solution. Check whether 1   y  1 x  1 c i.e. solution of the DE dx dy   0 can be defined explicitly or not? If it can be define then write down the explicit form? 1 x 1 y

05 Solution

Explicit solution: A solution of the differential equation of the form y= f (x) is called the explicit solution of the differential equation or roughly speaking if we can separate the dependent an independent variable in the solution then we say solution is explicit. For example dy / dt = y2-1/x has the solution y = 3+x2/3x2. It is explicit solution because dependent and independent variable are separate here of it is form of y = f(x). Implicit solution: A solution of the differential equation of the form G (x ,y)=0 is called the implicit solution of the differential equation or roughly speaking if you cannot separate the dependent an independent variable then that solution is said to be implicit. For example, dy/dt = 1+1/y2 has the solution y – tan-1 (y) = t+c. it is implicit solution because you cannot separate the dependent and independent variable or it is of the form of G (x, y) =0. Solution can be defined explicitly as follows.

1  ( y  1)( x  1)c 1 ( x  1)c 1 y 1 ( x  1)c y 1 

y

1   x  1 c

 x  1 c

Question 3

(b) Whether following differential equation is separable or not if it is separable then solve it?

1 (r  5r    5)dr   r 2  20r 2    20  d  0  2

dy 2 2 3 y t  t e dt

Solution

1 (r  5r    5)dr   r 2  20r 2    20  d  0 (r  5r    5)dr   r 2  20r 2    20  d  r   5   1  5   dr   r 2   20   1  20   d

 r  1  5 dr   r 2  1   20  d  r  1 dr    20  d   5  r 2  1   5  5  20  d dr  r 1   5   5 d   5  20  d dr  r  1   5    5 dr 25  d  d r 1  5 Integrate both sides dr 25  r  1   d     5 d ln r  1    25ln   5  c

10

dy 2 2 3 y t t e dt dy   t 2  t 2 e3 y  dt  0

 2

dy  t 2  e3 y  1 dt dy  t 2 dt e 1 e 3 y dy  t 2 dt 3 y 3y e  e  1 3y

3e 3 y dy  t 2 dt 3 y 3 1  e  1 3e 3 y dy   t 2 dt  3 y 3 1  e  ln(1  e 3 y ) t 3  c 3 3 3 y 3 ln(1  e )  t  3c (1  e 3 y )  et

3

 3c

(1  e 3 y )  et .e3c 3

(1  e 3 y )  c1et , say c1  e3c 3

Question 4

(a) Define a homogenous function. Check whether f  x, y  

x2 y is homogeneous or not explain. xy 3  x3 y

dy x2 y Also check whether the DE is solvable by method of homogenous equation or not if  dx ( xy 3  x3 y ) not then give reason? 05 Solution Homogeneous Function: A function f  x, y  is said to be homogeneous if f  tx, ty   t n f  x, y  , for some real number n e.g. f  x, y  =

2x 2 is homogeneous function 3xy+y2

but f  x, y  

x2 y xy 3  x 3 y

f  tx, ty  

t 3 x2 y t 4  xy 3  x3 y 

f  tx, ty  

x2 y t  xy 3  x 3 y 

f  tx, ty   t 1 f  x, y  This is homogeneous function dy x2 y Now to solve we can’t apply here the homogeneous method  dx ( xy 3  x3 y ) To solve the homogeneous differential equation

dy  f ( x, y) dx We use the substitution

v

y x

If f ( x, y) is homogeneous of degree zero, which we don’t have there.

Question 4

(b) Solve the following differential equation. dy 2 x  y  1  0 dx 2 y  x  1

Solution dy 2 x  y  1  0 dx 2 y  x  1

10

now put the values x  X  h, y  Y  k the given equation reduces to dy 2X  Y+2h  k+1  0 dx 2Y  X+2k  h  1 We choose h and k such that 2h  k+1=0;2k  h  1=0  h+2k  1=0 multiply this with 2 we get 2h+4k-2=0 then adding both 2h  k + 1=0 2h+4k  2=0 3k  1  0 k=1 3  2h  k+1=0  2h  1 3  1  0  2h+ 2 3  0  2h= 2 3  h= 1 3 so ultimately we get h=  1 3,k=1 3 dy 2X  Y = dx X  2Y This is a homogenous equation. We substitute Y=VX to obtain dY dV  X V dX dX by replacing all value s in the given DE dV dX dV dX dV dX dV dX

2X  VX X  2VX 2 V X V 1  2V 2  V  V 1  2V  X 1  2V 2  V  V 1  2V  X 1  2V X V 

dV 2  V  V  2V 2 X dX 1  2V dV 2  2V  2V 2 X dX 1  2V 1  2V 2 dV  dX 2 1V V X 2V  1 2 dV  dX  0 2 1V V X Integrating both sides 2V  1

2 dX  0 X ln | V 2  V  1| 2 ln | X | ln | c |

 1V V

2

dV  

X 2 (V 2  V  1)  c now replacing V=Y/X we will get the  Y2 Y  X 2  2   1  c X X  (Y 2  YX  X 2 )  c now replace the values of X,Y x=X  1 3, y  Y  1 3  X=x+1/3,Y=y-1/3  y  1/32   y  1/3  x+1/3    x+1/32   c  

 3 y  1   3x  1 3 y  1   3x  1  9c 3 y 2  6 y  1   3 xy  3 y  3 x  1  3x 2  6 x  1  9c 2

2

3 y 2  3x 2  2  3 xy  3 y  6 y  3 x  6 x  1  9c 3 y 2  3x 2  3 xy  9 y  9 x  3  9c y 2  x 2  xy  3 y  3 x  1  3c

Assignment 2 (Spring 2005) Maximum Marks 60 Due Date 20, April 2005 Assignment Weight age 2%

Question 1

(a) What is the difference between a linear equation (first order) and Bernoulli equation gives at least two examples of each.

05

Solution LINEAR DIFFERENTIAL EQUATION (FIRST ORDER) dy A linear differential Equation is an Equation of the form a  x   b  x  y  c( x) it’s more frequently dx dy used form  f ( x) y  g ( x) where f(x) and g(x) are both continuous functions of x. e.g. dx dy  x3 y  x 2 dx dy  xy  x dx BERNOULLI DIFFEREMNTIAL EQUATION Now the differential equation of the form

dy  f  x  y  g  x  y n where n is any real no is called dx

Bernoulli equation. e.g. dy 1  y  xy 2 dx x dy 1 x y 2 dx y DIFFERENCE The difference is that linear equation is a special type of Bernoulli’s equation when n is equal to 0 and 1 simply for n=0, 1 the Bernoulli’s equation becomes a linear equation. Question 1

(b) Solve the following linear differential equation y  x  2 y  dy dx 3

10

Solution y  x  2 y  dy dx 3

dy y  dx  x  2 y 3 

It is clearly not a linear equation it can be changed to a linear as in first order linear equation independent and dependent variables can be changed so writing equation as. dx x  2 y 3  dy y dx 1  x  2 y2 dy y nowit is linear in " x " p  y  1 y ,Q  y  2 1

IF  e

  y dy 1 ln    y

1 y Multiplying by 1/y on both sides we get IF  e

1 dx 1  x  2y y dy y 2 nowit is a differential of

d x   dx  y 

d x    2y dx  y  x  y 2  c  x  y ( y 2  c) y

Question 2

(a) Solve the following initial value problem and mention the type of the equation dy y x   3 dx 2 x y

Solution

, y(1)  2

10

dy y x   3 dx 2 x y It is a bernaoulli eqution in " y " where n  3 p  x  1 2x , Q  x   x dy y 4  x dx 2 x put t  y 4

y3

dt dy 1 dt dy  4 y3   4 y3 dx dx 4 dx dx Now the equation becomes 1 dt 1  tx 4 dx 2 x dt 2  t  4x dx x now IF will be IF  e

2

 xdx

 e 2ln x  eln x  x 2 2

Now multiplying both sides x 2 dt  2 xt  4 x3 dx now It is the differential of d tx 2   dt d (tx 2 )  4 x 3 dx x2

tx 2   4 x 3dx tx 2  x 4  c y 4 x2  x4  c By applying initial conditions Now put y=2, x=1 4 2 4  2  1  1  c

 c  15 So we get y 4 x2  x4  15 Question 2

(b)Show that y 2  4cx  4c2 is self orthogonal

Note: A family of curves is said to be self orthogonal if the orthogonal trajectories of the family is same as the DE of that family. 10 Solution y 2  4cx  4c 2    (1) now dy y dy  4c  c  dx 2 dx substituting this value in equation (1)

2y

we get  y dy   y dy  y  4x    4   2 dx   2 dx 

2

2

 dy   dy  y2  2x  y    y   dx   dx  so DE of given faimly is

2

2

dy  dy  y  2 xy   y       2  dx  dx  To get the family of orthogonal trajectories dy dy Replace by  1 dx dx 2  dy   dy  2 y  x2y    y   dx   dx  by replacing we get 2

  dy    dy  y 2  x  2 y  1   y  dx    dx     dy  2  dy  y  2 xy 1  y    dx   dx 

2

2

2

2

 dy   dy  y 2    2 xy    y 2  dx   dx  2

 dy   dy  2  y   2 xy    y     3 dx dx    

If we compare both differential equations (2) and (3) then came to know that both are the same So the given family of equation is self orthogonal Question 3

(a)Determine either the following curves are orthogonal to each other or not? If not then find the condition of orthognality. y 2  x 2  1 , x3  y  3 Note: Two functions are said to be orthogonal if product of the slopes is -1. Solution TO check both curves orthogonal or not, First of all differentiating both the equations w.r.t x y 2  x2  1 differentiatin wrt x dy  2x  0 dx dy 2 x  dx 2 y Now we check at ( x1 , y1 ) x dy m1   1 dx |( x1 , y1 ) y1 Now considering the second equation y  x3  3 2y

dy  3x 2 dx now at ( x1 , y1 ) dy  3 x12 dx |( x1 , y1 ) Now by applying condition of orthogonal m1m2  1 It is not satisfied Now both the curves are not orthogonal To find the condition we may proceed as follow and condition for orthognality is m1m2  1  0 m2 

putting the values x1 (3 x12 )  1  0 y1 3 x13 1  0 y1 3 x13  y1  0 y1  3 x13  0 This is the required condition.

07

Question 3

(b) Solve the following DE by the proper substitution xye y

2

dy y2 sinx +e = dx x

08

Solution xye y

2

dy y2 sinx +e = dx x 2

put e y =u dy du = dx dx 2 dy 1 du ye y = dx 2 dx x du sinx +u= 2 dx x du 2 sinx + u=2 2 dx x x 2ye y

2

Here p(x)=2/x And Q(x)=

sinx x2

2 I.F =exp(  dx)=x 2 x du x2 +2xu=2sinx dx d 2 (x u)=2sinx dx Integrate x 2 u=  2cosx+c x 2 e y =  2cosx+c 2

Question 4

(a) The population of a town grows at rate proportional to the population at any time. Its initial population of 50 decreased by 9% in 15 years what will be the population in 25 years? (Just make the model of the population dynamics as well as just describe the given conditions do not solve further) 03

Solution Suppose that P0 is the initial population of the town, as given P0 is 50 and P (t) the population at any time t. Then population growth can be represented by dP P dt dP  kP dt dP  kdt P Integrate both sides

ln P  kt  c P  ekt c P  e kt ec P  P0ekt

say P0  ec

P0 is the initial population of the town P0 =50 = P (0)

P (15) =50-

9  50  100

P (25) =? Question 4 (b) If a small metal bar whose initial temperature 30C is dropped in to a container of boiling water, how long will it take for the bar to reach 90C if it is known that its temperature increased 2C in a second? (Just make the model as well as just describe the given conditions do not solve further)

04 Solution Small metal bar has the initial temperature T (0) =30 C=T0 when dropped into a container of boiling water (temperature of surrounding matters a lot on the small metal bar) so suppose Tm is the temperature of the boiling water. As we know the boiling temperature of the water is 100c so Tm  100 dT Change in the temperature can be represented as and it is proportional to T  Tm  i.e. dt dT propotional to T  100 dt With conditions T (1) =32as 30+2and T (0) =30

We have too find the time when T=90 dT  k (T  100) dt dT  kT  100k dt Question 4

(c)The initially there were 50 milligrams of the radioactive substance present after 5 hours the mass increased by 4%. If the rate of decay is proportional to the amount of the substance present at any time, determine the half- life of the radioactive substance. (Just make the model as well as just describe the given conditions do not solve further) 03 Solution Suppose, radioactive substances = R Then its model would be dR proportional to R dt dR =kR dt Where k is the constant of proportionality With the initial conditions R (0) = 50, R (5) = 52 Since it is given after 5 hours radioactive substances Increased by 4%, thus 4% of 50 is 2 milligram That means 52, as we subtract 50+2=52 dR Then =kdt R After integration, we get lnR = kt + c R =e kt + c R =R0ekt where R0  e c Now we have to find half-life of radioactive Substances, that means t is required if R = R/2 i.e. R=25

Assignment 3 (Spring 2005) Maximum Marks 60 Due Date 02, May 2005 Assignment Weight age 2%

Question 1

(a) Define uniqueness of solution and condition of its existence? If y1 , y2 , , yn are n solutions, on an interval I , of the homogeneous linear nth-order differential equation dny d n1 y dy a n x  n  a n1 x  n1    a1 x   a0 x  y  0 on interval I are linearly independent then dx dx dx is it necessary w  y1 , y2 , , yn  must be non-zero? 05 Solution Let an ( x), an 1 ( x),..., a1 ( x), a0 ( x) and g (x) be continuous on an interval I and let an ( x)  0,  x  I . If x  x0  I , then a solution y(x) of the initial-value problem exist on I and is unique NOT PART OF THE SOLUTION: Always remember that solution of the differential equation differ by a constant as I told in the first assignment’s solution we have initial or boundary conditions to determine these unknown constants in the differential equations Secondly, If y1 , y2 ,

, yn are n solutions, on an interval I , of the homogeneous linear nth-order

dny d n1 y dy    a x    a1 x   a0 x  y  0 on interval I are linearly n 1 n n 1 dx dx dx independent then it is necessary that W ( y1 , y2 , , yn ) must be non-zero since we know

differential equation a n x 

Linear Independence of Solutions: The solutions y1 , y 2 ,, y n are linearly dependent if and only if W  y1, y 2 ,, y n   0, x  I

Question 1 (b) Verify that the given two-parameter family of functions is the complimentary solution of the nonhomogeneous differential equation y "' 5 y " 6 y '  2sin x  8; y  c1  c2e2 x  c3e3 x 05 Solution

Consider the associated homogenous differential equation y "' 5 y " 6 y '  0 ----------------------- (1) Given function is y  c1  c2e2 x  c3e3 x Taking derivatives of the function y '  2c2 e2 x  3c3e3 x

y "  4c2 e2 x  9c3e3 x y "'  8c2 e 2 x  27c3e3 x Putting all these in the given associated homogenous differential equation LHS  y "' 5 y " 6 y '  8c2e 2 x  27c3e3 x  5(4c2e 2 x  9c3e3 x )  6(2c2e 2 x  3c3e3 x )  8c2e 2 x  27c3e3 x  20c2e 2 x  45c3e3 x  12c2e 2 x  18c3e3 x 0 Hence y "' 5 y " 6 y '  0 is the complementary solution of the given non-homogeneous differential equation. Question 2

(a) Define linearly independence and linearly dependence of the functions? Comment either the functions given below are linearly independent or dependent f1 ( x)  xe x , f 2 ( x)  x 2e x , f3 ( x)  x3e x Solution Linear Dependence A set of functions

 f1 ( x), f 2 ( x),, f n ( x) is said to be linearly dependent on an interval I if  constants c1 , c2 ,, cn not all zero, such that c1 f1 ( x)  c2 f 2 ( x)  .  cn f n ( x)  0,

 xI

Linear Independence A set of functions

 f1 ( x), f 2 ( x),, f n ( x) is said to be linearly independent on an interval I if c1 f1 ( x)  c2 f 2 ( x)    cn f n ( x)  0,

only when

c1  c2    cn  0.

xI ,

10

f1 ( x)  xe x , f 2 ( x)  x 2e x , f3 ( x)  x3e x , First of all finding the derivatives of the function

f1 ( x)  xe  x f '1 ( x)  e  x  xe  x  1  x  e  x f ''1 ( x)  e  x  1  x  e  x   x  2  e  x f 2 ( x)  x 2e  x f '2 ( x)  2 xe  x  x 2 e  x   2  x  xe  x

f ''2 ( x)   xe  x   2  x  e  x   2  x  xe  x   x 2  4 x  2  e  x f 3 ( x )  x 3e  x f '3 ( x)  3 x 2 e  x  x 3e  x   3  x  x 2e  x

f ''3 ( x)   x 2 e  x   3  x  2 xe  x   3  x  x 2 e  x  xe  x  x 2  6 x  6  Now w  xe  x , x 2 e  x , x 3e  x  xe  x

x 2e x

x 3e  x

 1  x  e  x

 2  x  xe x

 3  x  x 2e  x

x

( x  2)e  x

2

 4 x  2  e x

x

x2

2  x x

3  x  x2

1  xe  x e  x e  x 1  x  ( x  2)

( x  2)

( x  2)

2

 4x  2 x  x2  6x  6 x2

2  x x

3  x  x2

x

2

 4x  2 x  x2  6x  6 x

x2

2  x x

3  x  x2

1  xe 3 x 1  x 

x x

1  xe 3 x 1  x 

xe  x  x 2  6 x  6 

x

2

 4x  2 x  x2  6x  6

 2  x x 3  x  x2 1  x   2  x 2 ( x  2)   x  4x  2 x  x  6x  6 3 x  xe   2 1  x  2  x x  x ( x  2)  x 2  4 x  2  

3  x  x2

  x  x2  6x  6     

 xe 3 x  x 2  x 2  4 x  6   x 2  2 x 2  6 x  6   x 2  9 x 2  10 x  2    x 3e 3 x  8 x 2  8 x  2   0

As Wronskian is not equal to zero so all function are linearly independent Question 2

(b) If functions f1 ( x), f 2 ( x),, f n ( x) possess at least n-1 derivatives on an interval I are linearly independent then is it necessary functions having Wronskian non-zero? If not then give the example (search a set of linearly independent functions such as their Wronskian is zero). 05 Solution If functions f1 ( x), f 2 ( x),, f n ( x) possess at least n-1 derivatives on an interval I are linearly independent then it is not necessary that functions having Wronskian non-zero for example f1 ( x)  0, f 2 ( x)  x, f3 ( x)  e x at least posses 2 derivatives on interval I

f1 ( x)  0, f 2 ( x)  x, f3 ( x)  e x f ''1 ( x)  0, f '2 ( x)  1, f '3 ( x)  e x f ''1 ( x)  0, f ''2 ( x)  0, f ''3 ( x)  e x Then it can easily notice that all above functions are linearly independent since Firstly, no function is the constant multiple of the other function Secondly, no one can be written linear combination of remaining two. Thus they all are linearly independent. 0 x ex But w  0, x, e x   0 1 e x  0 0 0 ex

This assures us if Wronskian of function is nonzero then there is no doubt set of functions will always be linearly independent but when Wronskian of functions is zero then set of function may be linearly dependent or may be linear independent . Question 2

(c) Find the 2nd solution of each of Differential equations by reducing order or by using the formula. x2 y '' 6 xy ' 4 y  0 ; y1  x 4  x3

Solution For solving this we use the formula  p ( x ) dx e y 2  x   y1  x   2 dx y1 ( x) x 2 y '' 6 xy ' 4 y  0 now dividing by x 2 on both sides we get y ''

6 y' 4 y  2 0 x x

05

now as y1 is given and it is given as y1  x 4  x3 now for sec ond solution we proceed as follows 6

y2  ( x 4  x 3 )   ( x 4  x3 ) 

e

 x dx

( x 4  x3 )2

dx

e6ln| x| dx ( x 4  x3 )2

x6 dx x 6 ( x  1) 2 1  ( x 4  x3 )  dx ( x  1) 2  ( x 4  x3 ) 

( x  1) 1 4 ( x  x3 ) 1 ( x 4  x3 )  ( x  1) 

x 3 ( x  1)  ( x  1)   x3 So this is the required second solution. Question 3 (a) Define fundamental solution set of a differential equation and what is criterion for existence of fundamental solution set of equation 05

Solution Any set y1, y2, y3,....... yn of n linearly independent solutions of the homogenous linear differential equation on an interval are said to be a fundamental set of solutions on the interval. Criteria for the existence of the fundamental solution set of for a linear nth order homogenous differential equation is, There always exists a fundamental set of solution for a linear nth order homogenous differential dny d n1 y dy equation. an ( x) n  an1 ( x) n1  ...  a1 ( x)  ao ( x) y  0 dx dx dx on the interval I Question 3

(b) Check weather x , x2 , x2 ln x in the interval (0,  ) form the fundamental set of solution the differential equation y "' 6 x2 y  4 xy ' 4 y  0 if it is fundamental solution what will be general solution. 10

Solution Now for fundamental solution set of y "' 6 x2 y  4 xy ' 4 y  0 differential equation we have to show that all the three functions are linearly independent. nowlet y1 , y2 , y3 are the three functions y1  x

y1'  1

y2  x 2

y1"  0

y2 '   2 x 3

y2"  6 x 4

y3  x 2 ln x

y3'   2 x 3 ln x  x 3

y3  x 2 ln x

y3'  x 3 1  2 ln x 

w  x, x 2 , x 2 ln x  x 2

x 2 ln x

1

2 x 3

x 3 1  2 ln x 

0

6 x 4

x

2

x x 1

y3"  x 4  6 ln x  5 

x 4  6 ln x  5 

x 2

y3"  6 x 4 ln x  2 x 4  3x 4

1 2 x

ln x

1  2 ln x  by taking common x 2and x 2 fromc2 , c3 respectively x 2  6 ln x  5 

1

x

6 x 2

0  2 x 1 x  2  6 x 4

1

x 1 1  2 ln x 

x 2  6 ln x  5 

 2  x 4  x 1 x 2 6 

1

1

x 1 1  2 ln x 

0 x 2  6 ln x  5 

1  2 ln x  1 1 x 0  6 ln x  5 

 ln x

2 x 1   0 6 x 2 

1

1 1  2 ln x   2 x 1 ln x x  6 ln x  5  0 1

1   3 x 1 

 x 4  x 3  12 ln x  10  6  12 ln x   x 2  6 ln x  5   6 x 2 ln x   x 4  4 x 3  5 x 2   x 6  4 x 1  5  0

As not equal to zero so these are independent and so forms a fundamental solution set. So general solution may be written as Y  c1 x  c2 x2  c3 x 2 ln x Question 4

(a) Find the solution of the following DE. y ''' 9 y '' 26 y ' 24  0; y  0  1, y '  0   2, y ''  0   3

Solution To solve this equation

10

y ''' 9 y '' 26 y ' 24  0 Suppose y  e mx

y '  me mx y ''  m 2 e mx y '''  m3e mx Substituting in equation m3emx  9m2emx  26memx  24  0 emx (m3  9m2  26m  24)  0 emx  0

or (m3  9m 2  26m  24)  0

m  2 is the root of equation then by use of synthatic division we get the roots as 2 1 9 26 24 2

14 24

1 7 12 0

 m  2   m2  7m  12   0  m  2   m2  3m  4m  12   0  m  2   m  m  3  4  m  3  0  m  2  m  3 m  4   0 general solution may be written as y  c1e 2 x  c2e3 x  c3e 4 x

applying y  0   1

1  c1  c2  c3    1 y '  2c1e 2 x  3c2e3 x  4c3e 4 x applying y '  0   2 2  2c1  3c2  4c3     2  y ''  4c1e 2 x  9c2e3 x  16c3e 4 x

applying y '  0   3

3  4c1  9c2  16c3     3

 EQ  2   2  EQ  1 2  2c1  3c2  4c3 2  2c1  2 c2  2c3 0  c2  2c3     4 

 EQ  3  4  EQ 1

3  4c1  9c2  16c3 4  4c1  4 c2  4c3 1  5c2  12c3  EQ  5  5  EQ  4  1  5c2  12c3 0  5c2  10c3 1  2c3  1 2  c3 then EQ  4 implies 0  c2 2 2  c2  2 2  c2  1 then EQ  1implies 1  c1  c2  c3  1  c1  1  1/ 2  c1  1/ 2 1 1  y   e 2 x  e3 x  e 4 x  2  2 Question 4 1 3

(b) The roots of an auxiliary equation are m1   , m2  2  i, m3  2  i . What is the corresponding differential equation? Solution 1 m2  2  i m3  2  i 3 Given that 1  m   0 (m  2)  i  0 (m  2)  i  0 3 The corresponding auxiliary equation will be, 1   m   ( (m  2)  i )( (m  2)  i )  0 3  1  2  m   (m  2)(m  2)  i (m  2)  i (m  2)  i   0 3   m1  

1 2   m   m  2m  2m  4  im  2i  im  2i  1  0 3  1 2   m   m  4m  5  0 3  m2 4 5  m  0 3 3 3 3 2 2 3m  12m  15m  m  4m  5  0 m3  4m 2  5m 

3m3  11m 2  11m  5  0 Finally the corresponding differential equation will be,

05

 3D3  11D 2  11D  5  0 

3

d3y d2y dy  11  11  5  0 3 2 dx dx dx

Assignment 4 (Spring 2005) Maximum Marks 40 Due Date 12, May 2005 Assignment Weight age 2%

Question 1 Solve the differential equation r ///  6r //  1  cos  .find complimentary rc and particular solution ( rp ) by the undetermined coefficient (superposition approach). 10 Solution r ///  6r //  1  cos  Complementary function To find rc , we solve the associated homogeneous differential equation r ///  6r //  0 Put r  en , r '  nen , r''  n2en , r '''  n3en Substitute in the given differential equation to obtain the auxiliary equation r ///  6r //  0

n 3  6n 2  0  n 2  n  6   0 n  0, 0, 6 Hence, the auxiliary equation has complex roots. Hence the complementary function is r  c  c   c e6 c

1

2

Particular Integral Corresponding to g ( )   cos  : Corresponding to f    1 :

3

rp1  A cos   B sin  rp2  C

Therefore, the normal assumption for the particular solution is rp  rp1  rp2 rp1  A cos   B sin  +C Clearly there is duplication of (i) The constant function between rc and rp2 .

To remove this duplication, we multiply rp2 with  2 . This duplication can’t be removed by multiplying with  . Hence, the correct assumption for the particular solution rp2 is

rp  A cos   B sin   C 2

r ' p   A sin   B cos   2C

Then

r '' p   A cos   B sin   2C r ''' p  A sin   B cos 

r ''' p1  6r '' p1  A sin   B cos   6   A cos   B sin   2C 

Therefore

r ''' p1  6r '' p1  sin   A  6 B   cos   6 A  B   12C

Substituting the derivatives of rp in the given differential equation and grouping the like terms, we have sin   A  6B   cos  6 A  B   12C  1  cos 

A  6B  0, 6 A  B  1, 12C  1  C  1 12 Solving these equations, we obtain A  6 B  6  6 B   B  1  37 B  1  B  1 37  A  6 37 A particular solution of the equation is rp  1 37  6cos   sin    1 12 2 So r  rc  rp r  c1  c2  c3e6  1 37  6cos   sin     2 12 Question 2 Factor the given differential operator also find the function annihilated by this D4  64D Solution D 4  64 D  D  D 3  64 

D 4  64 D  D D 3   4 

3

D 4  64 D  D  D  4   D 2  4 D  16  Since

Dn y  0

Annihilates each of the functions 1, x, x 2 ,

, x n 1

D Annihilates 1

Since ( D   ) n Annihilates each of the functions e x , xe x , x 2 e x ,  , x n 1e x ( D  4) Annihilates e Since differential operator -4 x

D 2  2D   2   2  n

is the annihilator operator of the functions

10

e x cos  x, xe x cos  x, x2e x cos  x, e x sin  x, xe x sin  x, x2e x sin  x,

, xn 1e x cos  x , xn1e x sin  x

2  4    2

 2   2  16   2    2  16   2  16  4   2  12    2 3 2

  2,   2 3

D

2

 4 D  16  Annihilates e2 x cos 2 3, e2 x sin 2 3

Therefore, the operator D4  64D  D  D  3  D2  4D  16  annihilates linear combination 1, e

-4 x

,

e2 x cos 2 3, e2 x sin 2 3

Question 3 Find the annihilator operator of the function e  2 e   2e Solution Suppose that

10

e  2 e   2e

y1( )  e , y2 ( )  2 e , y3 ( )   2e Then

 D  1 y1

 D  1

2

y2 

 D  1

3

y2 

e   0  D  1  2 e   0

 D  1

2

3  D  1   2e   0

Therefore, the product of two operators 3  D  1 Annihilates the given function

f ( )  e  2 e   2e

Question 4 Solve the differential equation y //  y  x2e x  5 .find complimentary yc and particular solution ( y p ) by the undetermined coefficient (annihilator approach). 10 Solution Step 1 The given differential equation can be written as ( D2  1) y  x2e x  5 Step 2 The associated homogeneous differential equation is

( D 2  1) y  0

 D  1 D  1  0 Roots of the auxiliary equation are complex m  1, 1 Therefore, the complementary function is yc  c1e x  c2e x

 D 1 x2e x  0, D  5  0 3 Therefore the operators  D  1 and D annihilate the functions 3

Step 3 Since

x 2e x and 5 . We apply D  D  1 to

the non-homogeneous differential equation D  D  1 ( D2  1) y  0 . This is a homogeneous differential equation of order 6. 3

Step 4 The auxiliary equation of this differential equation is 3 m  m  1 (m2  1) y  0  0

 m  0, 1,1,1,1, 1 Therefore, the general solution of this equation must be

y  c1  c2e x  c3xe x  c4 x2e x  c5 x3e x  c6e x

Step 5 Since the following terms are already present in y c c2e x  c6e x Thus we remove these terms. The remaining ones are c1  c3xe x  c4 x2e x  c5 x3e x

Step 6 The basic form of the particular solution of the equation is

y p  c1  c3xe x  c4 x2e x  c5 x3e x The constants c1 , c3 , c4 and c5 have been replaced with A, B and C . Step 7 Since y p  A  Bxe x  Cx 2e x  Dx3e x

y p  A   Bx  Cx 2  Dx3  e x Then taking derivative y ' p   Bx  Cx 2  Dx3  e x   B  2Cx  3Dx 2  e x y ' p   Bx  Cx 2  Dx3  B  2Cx  3Dx 2  e x y ' p   B  2C  x   C  3D  x 2  Dx3  B  e x Again taking derivative

3

y '' p   B  2C  2  C  3D  x  3Dx 2  e x   B  2C  x   C  3D  x 2  Dx 3  B  e x y '' p   B  2C  2  C  3D  x  3Dx 2   B  2C  x   C  3D  x 2  Dx 3  B  e x y '' p   2 B  2C   2C  6 D  B  2C  x   3D  C  3D  x 2  Dx3  e x Therefore

yp  y p   2 B  2C   2C  6 D  B  2C  x   3D  C  3D  x 2  Dx 3  e x  A   Bx  Cx 2  Dx 3  e x yp  y p   2 B  2C   2C  6 D  B  2C  x   3D  C  3D  x 2  Dx 3   Bx  Cx 2  Dx 3   e x  A yp  y p   2 B  2C   2C  6 D  B  2C  B  x   3D  C  3D  C  x 2  Dx 3  Dx 3  e x  A yp  y p   2 B  2C   4C  6 D  x  6 Dx 2  e x  A Substituting in the given differential equation, we have yp  y p  x 2e x  5

 2 B  2C   4C  6 D  x  6 Dx 2  e x  A  x 2e x  5 Equating coefficients of e x , x 2e x and the constant terms, we have 2 B  2C  0, 4C  6 D  0, 6 D  1,  A  5

Thus

B  C ,

2C  3D,

1 B , 4

1 C , 4

1 D , 6 1 D , 6

A  5 A  5

1 1  1 y p  5   x  x 2  x3  e x 4 6  4 ex y p  5   3x  3 x 2  2 x3  12

Step 8 Hence, the general solution of the given differential equation is y  yc  y p Or

y  c1e  c2e x

x

 5  3x  3 x  2 x 2

3

ex 12

Assignment 5 (Spring 2005) Maximum Marks 40 Due Date 21, June 2005 Assignment Weight age 2%

Question 1 State in words a possible physical interpretation of the given initial-value problems 1 d 2x  5x  0 (a) 32 dt 2 x(0)  2, x '(0)  1 1 d 2x  2.5 x  0 (b) 2 dt 2 x(0)  0.5, x '(0)  0.25 1 d 2x  3.5 x  0 (c) 8 dt 2 x(0)  1.5, x '(0)  0 15 Solution (a)Comparing

1 d 2x d 2x 1 with m  kx  0 , we got m= , k=5  5 x  0 2 2 dt 32 32 dt

W=mg W=32/32=1 Weight of 1 lb attached to a spring, is released from a point 2 units below the equilibrium position with initial downward velocity and the spring constant is 5lb/ft. 1 d 2x d 2x 1 (b)Comparing with  2.5 x  0 m  kx  0 , we got m= , k=2.5 2 2 2 dt dt 2 W=mg W=32/2=16 Weight of 16lb attached to a spring, is released from a point 0.5 units above the equilibrium position with initial upward velocity and the spring constant is 2.5lb/ft. 1 d 2x d 2x 1  3.5 x  0 m  kx  0 , we got m= , k=3.5 (c)Comparing with 2 2 8 dt dt 8 W=mg W=32/8=4 Weight of 4lb attached to a spring, body starts from rest from point 1.5 units below the equilibrium, and spring constant is 3.5 lb/ft.

Question 2 A 24lbweight, attached to a spring, stretches it 4 in. find the equation of motion if the weight is released from the rest from a point 3 in above the equilibrium point also find the time period(period of oscillation) and frequency as well as amplitude. 10 Solution As we know general equation of simple harmonic motion (un-damped free motion) is

m

d 2x  kx ----- (1) dt 2

d 2x Or 2   2 x  0 dt So basically we needed m, k thus we can find For consistency of units with the engineering system, we make the following conversions 12 inches  1 foot  1 inches  4 inches 

So,

1 foot 12

4 1 foot= foot . 12 3

Stretch  s  4inc 

4 1 ft  ft 12 3

Further weight of the body is given to be W  24 lb

But

W  mg

Therefore

m

W 24 3   g 32 4

or

m

3 slugs. 4

Since

Stretch  s  4inc 

4 1 ft  ft 12 3

Therefore by Hook’s Law, we can write 1 24  k    k  72 lbs/ft 3

Equation 1 is becomes

3 d 2x  72 x 4 dt 2 d 2x 4  72   x 2 dt 3

or

d 2x  96 x  0 . dt 2

Since the initial displacement is 3 inches  1 x  0   , 4

3 1 ft  ft 12 4 x  0   0

The negative sign indicates that body starts from rest above the equilibrium point at the distance of 3 inch from equilibrium point or ¼ foot from equilibrium point d 2x  96 x  0 dt 2

Subject to

1 x  0   , 4

Putting

d 2x xe ,  m 2 e mt 2 dt

x  0   0

mt

We obtain the auxiliary equation m2  96  0

or

m  4 6i

The general solution of the equation is x  t   c1 cos 4 6t  c2 sin 4 6t

Now, we apply the initial conditions. x  0  

1 4

 c1.1  c2 .0  

1 4

1 4

Thus

c1  

So that

x t  

1 cos 4 6t  c2 sin 4 6t 4

Since x  t    6c1 sin 4 6t  4 6c2 cos 4 6t .

Therefore

x  0   0  0  4 6c2 .1  0 Thus

c2  0 . Hence, solution of the initial value problem is 1 x  t    cos 4 6t 4

Time period T

2

Comparing

d 2x d 2x 2 with  96 x  0 we have   x  0 dt 2 dt 2

 2  96    4 6 It implies T

2   4 6 2 6

Frequency f 

1  2 6   T 2 

Amplitude A  c12  c2 2 

 1 4

2

 02  1 4

Question 3 Physically interpret the following differential equations. 1 d 2x dx 2 x 0 2 (a) 16 dt dt x  0   0, x '  0   1.5

d 2x dx  4  4x  0 2 (b) dt dt x  0   1, x '  0   2.01 d 2 x dx   16 x  0 (c) dt 2 dt x  0   1, x '  0   0

15

Solution (a)Comparing the given differential equation

1 d 2x dx 2 x 0 2 16 dt dt With the general equation of the free damped motion

m

d 2x  dx   β    kx  0 2 dt  dt 

d 2 x β  dx  k    x  0 dt 2 m  dt  m d 2x dx  2λ   2 x  0 2 dt dt

So here

m =1/16, β =2, k =1 W=mg W=32/16=2lb we see that 2λ 

β 2 k 1  , 2    16 m 1 16 m 1 16

λ  16,

So that

4

λ2  2  0

System is Over-damped motion. The problem represents “A 2lb weight is attached to a spring whose spring constant is 1lb/ft i.e. k=1lb/ft. the system is Overdamped with resisting force numerically equal to 2 times of the instantaneous velocity i.e.  dx   dx  Damping force  -β    2   ”  dt   dt  Inspection of the boundary conditions:

x  0  0, x '  0   1.5 Reveals that the weight starts from equilibrium position with upward velocity of 1.5 ft/sec Solution (b)Comparing the given differential equation d 2x dx  4  4x  0 2 dt dt

With the general equation of the free damped motion

d 2x  dx  m 2  β    kx  0 dt  dt  d 2 x β  dx  k    x  0 dt 2 m  dt  m

d 2x dx  2λ   2 x  0 2 dt dt

So here

m =1, β =4, k =4 W=mg W=32=32lb we see that β 4 k 4  , 2    4 m 1 m 1 λ  2, 2

2λ 

So that

λ2   2  0

System is critically -damped motion. The problem represents “A 32lb weight is attached to a spring whose spring constant is 4lb/ft i.e. k=4lb/ft. the system is critically damped with resisting force numerically equal to 4 times of the instantaneous velocity i.e.  dx   dx  Damping force  -β    4   ”  dt   dt  Inspection of the boundary conditions:

x  0  1, x '  0   2.01 Reveals that the weight starts from1 unit above the equilibrium position with a downward velocity of 2.01 ft/sec Solution (c)Comparing the given differential equation

d 2 x dx   16 x  0 dt 2 dt With the general equation of the free damped motion

m

d 2x  dx   β    kx  0 2 dt  dt 

d 2 x β  dx  k    x  0 dt 2 m  dt  m d 2x dx  2λ   2 x  0 2 dt dt

So here

m =1, β =1, k =16 W=mg W=32=32lb we see that 2λ 

β 1 k 16  ,  2    16 m 1 m 1

1 λ , 2

So that

4

λ2   2  0

System under -damped motion The problem represents “A 32lb weight is attached to a spring whose spring constant is 16 lb/ft i.e. k=16 lb/ft. the system is under -damped with resisting force numerically equal to 1 times of the instantaneous velocity i.e.  dx   dx  Damping force  -β    1  ”  dt   dt  Inspection of the boundary conditions:

x  0   1, x '  0   0 Reveals that the weight starts from rest 1 unit below the equilibrium position

Question 4 (a) A 16-lb weight stretches a spring 8/3 ft. Initially the weight starts from rest 2-ft below the equilibrium position and the subsequent motion takes place in a medium that offers a damping force numerically equal to ½ the instantaneous velocity. Find the equation of motion, if the weight is driven by an external force equal to f t   10 cos3t. 10 (b)Solve the following differential equation. x2 y  2 xy  2 y  x3 ln x

10 Solution (a) Given w=16 lb and s=8/3 ft , β = ½ , f t   10 cos3t. Initially the weight starts from rest 2-ft below the equilibrium position and the subsequent motion takes place in a medium i.e x  0   2, x '  0   0 then differential equation of forced motion

d 2x  dx       kx  f  t  2 dt  dt  W 16 1 W  mg  m    g 32 2 F 16 F  ks  k   6 s 83 m

1 d 2 x 1 dx   6 x  10cos 3t 2 dt 2 2 dt d 2 x dx or   12 x  20cos 3t dt 2 dt First consider the associated homogeneous differential equation. d 2 x dx   12 x  0 dt 2 dt dx d 2x mt mt Put xe ,  me ,  m 2 e mt 2 dt dt Then the auxiliary equation is: m 2  m  12  0

1 1 m 2  m    12  0 4 4 1  48  1  0 m   2 4  2

2

1 47  m    2 4  1 47  i 2 2 1 47  m  i 2 2 Thus the auxiliary equation has complex roots 1 47 1 47 m1  m    i, m2  m    i 2 2 2 2 So that the complementary function of the equation is 47 47  t 2  x e c cos t  c sins t   1 c 2 2 2   To find a particular integral of non-homogeneous differential equation we use the undetermined coefficients, we assume that x p  t   A cos3t  B sin 3t m

Then

xp  t   3 A sin 4t  3B cos 4t xp  t   9 A cos 4t  9B sin 4t

So that

xp  xp  12 x p  9 A cos 4t  9 B sin 4t  3 A sin 4t  3B cos 4t  12 A cos3t  12 B sin 3t   3 A  3B  cos 4t   3 A  3B  sin 4t Substituting in the given non-homogeneous differential equation, we obtain 3A  3B  cos 4t   3 A  3B  sin 4t  20cos3t Equating coefficients, we have 3 A  3B  20 3 A  3B  0 Solving these equations, we obtain 10 10 A , B 3 3 10 Thus x p  t    cos3t  sin 3t  3 Hence the general solution of the differential equation is: 47 47  10 t 2  x t   e t  c sins t    cos 3t  sin 3t   c1 cos 2 2 2  3  1 t 2  47 47  x  t   x  t    e t  c sins t  c1 cos 2 2 2 2  

Now

47  t 2  47 47  10 e t  c cos t    3  sin 3t  cos 3t   c1 sin 2 2 2 2  3 

x  0   2 gives c1.1 

Also

10 10 4  2  c1  2    3 3 3

x0  0 gives 0

47 10 1 4 c2  3     2 3 2 3

47 2 c2  10  2 3 64 c2  3 47 0

or Hence the solution of the initial value problem is: 47 64 47  10  t 2  4 x t   e t sins t    cos 3t  sin 3t   cos 2 2  3 3 47  3

Solution (b) x 2 y  2 xy  2 y  x 3 ln x put y  x m , y  mx m 1 y  m  m  1 x m  2

 x m  m 2  m  2m  2   0

 m  2  m  1  0 m  1, 2 yc  c1 x  c2 x 2  y1  x, y2  x 2 By V .D.P  y p  u1 y1  u2 y2 W

x

x2

1

2x

 2x2  x2  x2 x2

W1 

0

W2 

x

3

x ln x 1

2x 0 3

x ln x

  x 5 ln x  x 4 ln x

W1   x3 ln x W On int egrating u1 

 u1 

 x4  1  ln x   4  4

W2  x 2 ln x W On int egrating u2 

 u2 

x3  1  ln x   3 3

 y p  u1 x  u2 x 2  x5  1  x5  1 ln x      ln x   4  4 3  3  ln x 1 ln x 1   y p  x5       3 9  4 16  yp 

 ln x 7  2  y  x 5     c1 x  c2 x  12 16  9  

Assignment 6 (Spring 2005) Maximum Marks 60 Due Date 05, July 2005 Assignment Weight age 2%

Question 1

(a) Define sequence and series define both with at least two examples, can every sequence be 

xn x 2 x3 represented as a series or not? Consider the series   1  x    ... find the derived 2! 3! n 0 n ! series by differentiating the series. Show that both have same radius of convergence. 10 NOTE: A series is said to be derived series which is obtained by integrating or differentiating term by term a given series. Solution Sequence: The sequence is defined as the function having the natural numbers as its domain N= {1, 2, 3…}. Series: The series is defined sum of the terms of a sequence. Examples n  1 2 3 { }n 1  , , ,...           (1) n2 3 4 5 2 n 1 4 9 { } , , ...           (2) n 1  2n  1 3 5 7 now forseries,seriesarethesumof thetermsof asequence 

 ar

n 1

 a  ar  ar 2  ar 3  ...

n 1 

1

1

1

1

 k (k  1)  1.2  2.3  3.4  ... k 1

Every sequence definitely can be represented as a series by simply adding the terms of that sequence. Both the sequence 1 and 2 can be represented as a series like  n 1 2 3     ...  3 4 5 n 1 n  2 

n2 1 4 9     ...  3 5 7 n 1 2n  1 Now we are going to find the derived series The given series is

xn x 2 x3  1  x    ...  2! 3! n 0 n ! nowto getderived serieswedifferentiatetermbyterm x 2 x3  ... 2 6 wewillgetthisserieswhichissameastheaboveseries

 0 1 x 

xn x 2 x3  1  x    ...  2! 3! n 0 n ! soboththeserieswillhavethesameradiusof convergencebut forisseries radiusof convergenceis lim n 

1  n  1! cn 1 n! n!  lim n   lim n   lim n  0 cn 1 n! n ! n  1  n  1!

Question 1 x3 x5 x 7   ... the find the few terms of the power series of Cscx 3 5 7 Show all the steps involved each step has its own importance.

(b)If Sinx  x 

Solution Co sec x  

1 Sinx   x x x7 x9   x-      9   3 5 7 3

5



 x3 x5 x7 x9    x-       9   3 5 7 

  x 2 x 4 x 6 x8   1-(       x 3 5 7 9 

05

I Now as we have that Sinx  x 

x3 x5 x 7   ... 3 5 7

pplyingtheboinomialtheorem i.e. (1+x)n =1+nX+n(n-1)/2! X2+n(n-1)(n-2)/3! X3... When n is a  ve integer or it is in the form of fraction then we use the following formula

.

  x 2 x 4 x 6 x 8    x 2 x 4 x 6 x 8   1+(-1)                5 7 9   5 7 9   1  3  3 =   x    x 2 x 4 x 6 x 8           ...    5 7 9   3   

  x 2 x 4 x 6 x8  x 2 x 4 x 6 x8 1+(-1)(      (        1 3 5 7 9  3 5 7 9    2 4 6 8 x x x x x     ( 3  5  7  9   ...    x 2 x 4 x 6 x8   x 4 x8   x6 x8 1+      neglecting  higher  tems       neglectinghigherterms        5 7 9 15 21 1  3   4 25     6 8   x x  x     neglectinghigherterms   ...  45  27    x 2 x 4 x 6 x8 x 4 x8 x6 x8  1            1 3 5 7 9 4 25 15 21    6 xx  x8     ...  45  27   x2 x4 x4 x6 x 6 x 6 x8 x8 x8  1           1 3 5 4 7 15 27 9 25 21    6 xx  x8     ...  45  27   1  x 2 9x 4 296x 6 206x 8  1-     ...  x 3 20 945 945 

So it is the required power series for the given function Question 1 

(c) Find the radius of convergence of

 (1)

n 1

n 1

. Solution 

compareit  (1) n 1

 cn  (1)

n 1

n 1

n( x  2) with cn ( x  x0 ) n n

n  cn1  (1)

n 1

n2

 n  1

n( x  2)n

05

R  lim n

cn 1 cn

R  lim n

(1) n  2 (n  1) (1) n 1 n

n 1 1 n So the radius of convergence is 1 R  lim n

Question 2 

(a) Find the function represented by the following power series

x

n

05

n0

Solution

now forn  0,1, 2,3... wecangettheseries 

x

n

 1  x  x 2  x 3  x 4 ...

n 1

soitisaninf initegeometricseriesso finding itsum a 1 r herea  1andr=x

S 

1 so  1-x Question 2

(b) Show that the given DE has a regular singular point at x=0.Determine the indicial equation, the recurrence relation and the roots of the indicial equation. 10 xy " y  0 Solution xy " y  0 ins tan dard  formitcanbewrittenas y 0 x nowbydefinationof sin gular po intx=0isasingularpoint y "

andp(x)=0forwhich x-0)isanalytic 1  andq(x)= forwhich x-0) 2 isanalytic x x

sothepointx=0isregularsingularpoint. xy " y  0              (1)

nowwesup posethereisasolutionof the form 

y   an x r  n n 0 

y '   an (r  n)x r  n 1 n 0 

y ''   an (r  n)(r  n  1) x r  n  2 n0

nowsubstitutiong allthesesvaluesinequationno1 

x  an (r  n)( r  n  1) x r  n  2   an x r  n  0 n 0

n 0

 a (r  n)(r  n  1) x n0

n

r  n 1

  an x r  n  0 n0

a0 r (r  1) x r 1   an (r  n)(r  n  1) x r  n 1   an x r  n  0 n 1

n 0

n0

n 1

nowwecanwrite an x r  n   an 1 x r  n 1 nowputtingthisvalueintheequation 

n 1

n 1

n 1

n 1

a0 r (r  1) x r 1   an (r  n)(r  n  1) x r  n 1   an 1 x r  n 1  0 a0 r (r  1) x r 1   an (r  n)(r  n  1) x r  n 1   an 1 x r  n 1  0 

a0 r (r  1) x r 1   [an (r  n)(r  n  1)  an 1 ]x r  n 1 n 1

nowwehave gotindicialequationasr (r  1)  0 r  0r  1

nowcomparingthecofficentof  x r  n 1 wegetan (r  n)(r  n  1)  an 1  0 an 1 (r  n)(r  n  1) so forthe greaterrootr  1 an 

wegettherelation an 

an 1 (1  n)(n)

nowreplacingvaluesof n  1, 2,3... a0 1.2 a a forn=2a2  1  0 2.3 2.3! a a forn  3a3  2  0 3.4 3.4! socontinuinginthismannerwewill get forn  1a1 

(1) n a0 n !(n  1)! so fortheseriessolutionwewillgetthe following seriessolution

an 

(1) n a0 x x2 y1 ( x)  x[1    ...   ...] 1!2! 2!3! n !(n  1)!

Question 3 (a) Define the ordinary and singular point. Is it necessary that ordinary and singular points are always real? Find the singular point for the following differential equations. (1) ( x2  4) y  2 xy  6 y  0  1   (2)   y  3xy  0  x 1 

08

Solution Ordinary point: A point x 0 is said to be a ordinary point of a differential equation a2 ( x) y   a1 ( x) y   a0 ( x) y  0 if a ( x) a ( x) both P(x) = 1 and) Q(x)= 0 are analytic at x 0 . a2 ( x) a2 ( x) Singular point: A point that is not an ordinary point is said to be singular point of the equation Or A point x 0 is said to be a singular point of a differential equation a2 ( x) y   a1 ( x) y   a0 ( x) y  0 if a ( x) a ( x) both P(x) = 1 and) Q(x)= 0 or either one of these are not analytic at x 0 . a2 ( x) a2 ( x) Both singular and ordinary points need not be real numbers (1) ( x2  4) y  2 xy  6 y  0 The equation ( x2  4) y  2 xy  6 y  0 has the singular points at the solutions of x2  4  0 , namely, x  2i . All other finite values, real or complex, are ordinary points.

 1   (2)   y  3xy  0  x 1   1   The equation   y  3xy  0 has no singular point. But having ordinary point at the solutions of  x 1  x  1  0 , namely, x  1 .

Question 3 (b) If x  x0 is an ordinary point of the differential equation y   P( x) y  Q( x) y  0 , can we always find two linearly independent solutions in the form of power series centered at x 0 : 

y   c n ( x  x0 ) n . n 0

Give the reason? Solution

02

Yes! We can always find the two linearly independent solutions and THEOREM (Existence of Power Series Solution) gives us authentication .i.e. “If x  x0 is an ordinary point of the differential equation y   P( x) y  Q( x) y  0 , we can always find two linearly independent solutions in the form of power series centered at x 0 : 

y   c n ( x  x0 ) n . n 0

A series solution converges at least for x  x0  R , where R is the distance from x 0 to the closest singular point (real or complex)” Question 4 (a) Define the regular singular and irregular singular point. Find the singular point for the following differential equations. (1) ( x2  4) y  2 xy  30 y  0 (2)  x  1 y  2  x  1 y  5 y  0

06

Solution Regular singular point A Singular point x  x 0 of the given equation a2 ( x) y  a1( x) y  a0 ( x) y  0 is said to be a regular a ( x) a ( x) singular point if both ( x  x0 ) P( x)  ( x  x0 ) 1 and ( x  x0 )2 Q( x)  ( x  x0 )2 0 are analytic a2 ( x) a2 ( x) at x 0 . Irregular singular point A singular point that is not regular is said to be an irregular singular point of the equation Or

A Singular point x  x 0 of the given equation a2 ( x) y  a1( x) y  a0 ( x) y  0 is said to be an irregular a ( x) a ( x) singular point if both ( x  x0 ) P( x)  ( x  x0 ) 1 and ( x  x0 )2 Q( x)  ( x  x0 )2 0 or either one a2 ( x) a2 ( x) of them are not analytic at x 0 . (1) ( x2  4) y  2 xy  30 y  0 are singular points of the equation x  2 2 Because x  4  0 or x  2 . Now write the equation in the form 2x

y 

or

P( x) 

y 

30

y0 ( x 2  4) x2  4 2x 30 y  y  y0 ( x  2)( x  2) ( x  2)( x  2)

2 x ( x  2)( x  2)

and

Q( x) 

30 ( x  2)( x  2)

Then Clearly x  2 are regular singular points. The factor ( x  2) appears at most to the first powers in the denominator of P(x) i.e.1 and at most to the second power in the denominator of Q( x)i.e.1 then x  2 is a regular singular point. Similarly, for x=-2. (2)  x  1 y  2  x  1 y  5 y  0 x  1 are singular point of the equation Because x  1  0 or x  1 . Now write the equation in the form

 x  1 y  5 y  0  x  1  x  1

y  2

or

P( x) 

y 

2

 x  1

0

2

 x  1

0

y 

and

5 y0  x  1 Q( x) 

5  x  1

Then Clearly x  1 are regular singular points. The factor ( x  1) appears at most to the first powers in the denominator of P(x) i.e.0 and at most to the second power in the denominator of Q( x)i.e.1 then x  1 is a regular singular point. Question 4

(b) If x  x 0 is a regular singular point of equation a2 ( x) y  a1( x) y  a0 ( x) y  0 , then is it necessary there exists always two series solution of the form if not then Give the reason?

02

Solution No it is not necessary there exist two solutions Frobenius’ Theorem gives us authentication that there exist just one power series solution .i.e. If x  x 0 is a regular singular point of equation a2 ( x) y  a1( x) y  a0 ( x) y  0 , then there exists at least one series solution of the form y  ( x  x0 )

r

 cn ( x  x0 )

n

n 0

  cn ( x  x0 )n  r n 0

where the number r is a constant that must be determined. The series will converge at least on some interval 0  x  x0  R. Question 4 (c) Define second order Bessel differential equation and Bessel function. Find J  3/ 2  x 

07

Solution Bessel differential equation: A second order linear differential equation of the form x2

d2y dy  x  x2  v2 y  0 2 dx dx

is called Bessel’s differential equation. Bessel function:

d2y dy  x  x 2  v 2 y  0 differential equation is usually denoted by J v x  and is 2 dx dx known as Bessel’s function.

Solution of the x 2

And now to find J 3/ 2  x  Consider

J v 1  x   J v 1  x   For v  

2v Jv  x  x

1 2

J  1 1  x   J  1 1  x   2 2

2 1 x

2 J 1

2

 x

J 3 J 3

2 2

 x J1  x   2

 x  

1 J 1  x x  2

1 J 1  x J1  x x  2 2

As given (we know) J1/ 2 ( x) 

J 3 

2

J 3

 x 

2

2 2 sin x, J  1/ 2 ( x)  cos x x x

 1 2 2 cos x   sin x  x x  x 

 x 

2  cos x   sin x   x  x 

Assignment 7 (Spring 2005) Maximum Marks 60 Due Date 13, July 2005 Assignment Weight age 2%

Question 1 (a) write the give DE as a system in normal form d4y d3y 3 4  3 3  6 y  10 dt dt Solution y  x1 , y '  x2 , y "  x3  y "'  x4

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nowx1'  y ' sox1'  y '  x2  x1'  x2 now y "  x2' sox2'  x3 asy "  x3 nowdifferentiating bothsides y "'  x3' sox3' x4 x4'  y "" 10  3 x4  2 x1 4 sotherequired normal  formis x4' 

x1'  x2 x2'  x3 x3'  x4 10  3 x4  2 x1 4 Question 1 (b)What is degenerate system Explain in your own words with example( no definition is required from books) Write the following system in the normal form if possible d 3 y d 3x dx dy 2 3  3  10t 2  4  6 dt dt dt dt 08 3 2 3 d x d y d y dy 2 3  5x  3 2  4 3  4 dt dt dt dt Solution x4' 

The system of DE is said to be degenerate system If it cannot be reduced to a linear system in the form of normal form .Some time the situation arises that when we write the highest derivative in the form of remaining variables then we try to eliminate the both the highest derivatives eliminate at same time so it is not possible to write it in the form of linear normal from this type of systems are known as degenerate systems. The solution of the given system is not possible it is a degenerate system because d 3 y d 3x dx dy 2 3  3  10t 2  4  6 dt dt dt dt 3 2 3 d x d y d y dy 2 3  5x  3 2  4 3  4 dt dt dt dt sowritinging again d 3 y d 3x dx dy  3  10t 2  4  6                 (1) 3 dt dt dt dt 3 3 2 d y d x d y dy 4 3  2 3  5 x  3 2  4              (2) dt dt dt dt multiplying (1)by2and by1  2

d3y d 3x dx dy  2  20t 2  8  12 3 3 dt dt dt dt 3 3 2 d y d x d y dy 4 3  2 3  5 x  3 2  4 dt dt dt dt 4

It is clear that it is a degenerate system Question 2 (a)Explain the difference between the echelon form and reduced echelon form.

05

Solution Explain the difference between the echelon form and reduced echelon form The difference in the reduced echelon and echelon form is that for each pivot element aij in the echelon form aij =1 and for each element in that column in which pivot exists for Which i is less than the pivot i are non zero and reaming all element are zero. Now in the case of reduced echelon form each pivot element aij =1 and all element in that column should be zero in which pivot exists, the best example of reduced echelon form is identity matrix. Question 2 (b)Solve the following system of equations by using gauss elimination method

6 x1  6 x2  6 x3  6 2 x1  4 x2  6 x3  12 10 x1  5 x2  5 x3  30 Solution The augmented Matrix may be written as

10

1/ 6 R1  6 666   1   2 4 612  R  2 3 1/ 2 R 2     10 5530    1  R 3  1  R  R1 R  1 4  2  R 3  2 R  1   0  R  R2 R  0 5  1 R  R 1  2 0  R 0  R 2 1  0  R 0  R 3  R2 1  0  R   R23    now by backward substitution we will get the values of all variables x1  x2  x3  11/ 5 Question 3

(a) Define are Eigen values and vectors. Solution The equation Ax  y can be viewed as a linear transformation that maps a given vector x into a new vector y .to find such a vector we set y=  x, where  is a scalar proportionality factor, and seek solutions of the equations Ax=  x (A-  I)x=0 ---------------(1) Where  satisfy the equation det (A-  I)=0and are called eigenvalues of matrix A And non zero solution of equation 1 obtained by using the Eigen values are known as Eigen vectors. 05 Question 3 (b) Find the Eigen values and vectors of the following  5 2 4  1 1 1     4 3 3 Solution

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 5 2 4  Let A  1 1 1  then Eigen values can be obtained as    4 3 3

The characteristic equation of the matrix A is

det  A   I  

5

2

1 4

1  3

4 1 0 3  

Expanding the determinant by the cofactors of the second row, we obtain

 5      1     3     3   2    3     4   4   3  4 1      0  5      3    3   2  3   2    3     4   4   3  4 1      0

 2     5     2   7     4   7  4   0 10  5  2       14  2     28  16   0 10  3       14  2     28  16   0 2

2

2

2

3

3

 3  3 2  24  42  0 have eigen values

1  4.4704 2  5.0697 3  1.6006

But its have complex Eigen vectors corresponding to Each Eigen value.

Question 4

Solve the following system of DE by the method of undetermined coefficients. dx  6 x  y  2et  1 dt dy  4 x  3 y  e t  5t  7 dt Solution

 6 1 X '  X  F (t )  4 3  dx   dt   x  2  0  1 X '    X   andF(t)=   e  t    t     dy   y 1  5   7     dt  now firstsolvethethehomogenoussystem  6 1 X'=  X  4 3 nowdet( A   I ) 

6

1

4

3

0

 (6   )(3   )  4  0   2  9  14  0  1  2,2  7 nowfor1  2 1   k1  0  6  2    4 3  2   k2  0    4 1  k1  0   4 1 k  0     2    4k1  k2  0   4k  k  0   1 2   sosolving weget k1  1,k2  4 1 K1     4  now for  7  1 1   k1  0   4 4  k  0     2    k1  k2  0 nowcho sin g arditraryvalues 1 K2    1 1  1 nowas X c  c1   e 2t  c2   e 7 t  4  1 

 2   0  1 nowF (t ) 1   e t    t    1  5   7  sothe particularsolutionis a  a  a  X p   3  e t   2  t   1   b2   b1   b3  asbyreplacing thee t inF(t)one 2t theisaneigenvaluesocorrectformof particularsolutionis a  a  a  a  X p   4  te 2t   3  e 2t   2  t   1   b4   b2   b1   b3 