Jee 2014 Booklet6 Hwt Solutions Differential Equations

Vidyamandir Classes

Solutions to Home Practice Test/Mathematics Differential Equation 1.

x

HWT - 1

dy dy y y2  y  x2  y 2    1 2 dx dx x x

Put y  mx 

x

dy dm x m dx dx



dm  m  m  1  m2 dx

dm  1  m2 dx

or

x

or

m  m 2  1  ec x

or

y  y 2  x 2  kx 2

or

ydy  xdx

or

dm 1 m

2



dx . x

Integrate both sides to get : log m  m 2  1  log x  c

2.

y 2  x2  ec x x

y  x

or

According to the question

dy .yx dx

Integrate both sides to get :

y 2 x2  c 2 2 As Curve passes through (5, 3)

or

c  8

Equation of curve is x 2  y 2  16

 3.

9 25  c 2 2

x 2  2 x  1 dy x2 y.  dx x  x  1 x 1 x2

 x x 1 dx Integrating factor  e   Now



x2 dx  x  x  1





1 3 1 x 1 1 3 x 1 log x 2  x   log  log x 2  x  log 1 2 2 x 2 2 x 2 2

 log

5.



1 3  2 x  1  dx 2 2 dx  1 log x 2  x  3 2 2 2 2 2 x x 1 1  x 2   2     

x2  x 

x3 2 x2  log x 1  x  1 x  1

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

 I.F. = e  Put So,

1 x

dx

 e 2



x2 x 1

x

x t  x

Integrating factor 

dy  y e 2 x   dx x x



General solution is y . e 2

ye 2



1 2 2

 2e

2 t

x





e 2

x

dx

x

dx  dt

dt   e 2t dt   e 2

x

C



y  C . e2

x

1

2

6.

dy dy  dy   p to get In the equation x     y  x   y  0 , put dx dx  dx 

xp 2   y  x  p  y  0

VMC/Differential Equation



p

 x  y    y  x 2  4  x   y  2x

18

HWT-Solutions/Mathematics

Vidyamandir Classes 

 x  y    x  y   1 or

y x

2x

dy  1 or dx



y x

dy  1 , we have y  x  c1 When dx dy  y  , we have xy  c2 dx x As curves pass through (2, 3) we have y = x + 1 and xy = 6

When

For another point of intersection we have x  x  1  6

1 dy 1 1  tan x   sec x . Substitute  v 2 dx y y y dv  tan x .v  sec x [L.D.E] dx



8.

9.



 x  3 x  2   0

 3x  1 dx  c

y

As

y  0   0, 0  0  0  c  c  0



tan 1  y   x3  x

dy 2

1



y  Ae3 x  Be5 x  d2y 2

8



1 dy dv  y dx dx



sec x  y  c  tan x 





dx

2

or tan 1  y   x3  x  c





y  tan x3  x 

or

tan x3  tan x 1  tan x3 . tan x

dy d2y  3 Ae3 x  5 Be5 x   9 Ae3 x  25 Be5 x dx dx 2

dy  15 y  0 dx 1

1

dy 1 ecos x dy 1 ecos x  y   . y dx dx 1  x2 1  x2 1  x2 1  x2

Now, Integrating factor (I.F.) = e

1



1 x 2

dx

 ecos

1 x

Differential Equation 2.(A)

x  3 , 2

or

  y  1

dy  3x 2 y 2  3x 2  y 2  1  3x 2  1 dx



10.

or

The other point of intersection is  3,  2 

 7.(C)

x2  x  6  0

or

xdy  ydx y

2

 xy 1  log x  dx

x  x 



x 1  log x  dx   y d  y      2

HWT - 2

x  d    xy 1  log x  dx  y  x2 y



2

2 x3  2   log x   c 3  3 

by parts

4.

dy x 2  y 2 1   y / x    dx 2 xy 2 y / x Put y = mx to get

2

dy dm m x dx dx

dm 1  m2 1 m dm 1 m 1  m2 .x    .x   dx 2m 2m 2 dx 2m 2 2m



m



 1 m 2m

2

dm 

 x c dx

As curve passes through (2, 1) 4 4   2k  4 1 3



k



So, the equation of curve is 3 x  2 x 2  y 2

VMC/Differential Equation



or  log 1  m 2  log  x   c



or

1 1 m

2



1 1

y

2



x2 x  y2 2

 kx

x2

2 3

19

HWT-Solutions/Mathematics

Vidyamandir Classes 5.(D)

Let y  tx

dy dt tx dx dx



   1   t dt

  1  t dt



2

t  

6.(B)

 y x y2 1

7.

dt

dx

t  

[ x y  log cx is general solution]



1

log 3  0  c



y1  3 x  1



y  x3  3 x  1



2





log y1  log x 2  1  c where





x cos

 1   x

 x   y2     2  y x

2x 2

x y xdx 1     t   t     x dx t  y

t  

1     t 2 t



1

y 

dy dx

3 x 0



c  log 3



y  x3  3x  c1 where y  0   1  c1  1

y y  y dx  x dy   y sin  x dy  y dx  x x y y . tan  x dy  y dx  x x



 y dx  x dy  



 y dx  xdy   tan  y  . d  y  x  

xy

x   

d  xy  xy

 y  y  tan   d    x  x

Integrate both sides to get :

log  xy   log sec 8.

It is order = 2, degree = 1

9.

x

2

y c x

or

xy  k sec

y x



 y 2 dy  xy dx



dy xy  dx x 2  y 2



t

Let y  tx

xdt t  dx t 2  1

t  1dt  2





As

y(1) = 1





 ln 1 

1



1  y  x2 n    2  nx  2  x  2y

t 3

2 1

x

dx



xdt t 3  2 dx t  1



 nt 



c

1  2

x2 2e



ye

1

 ln x  c

2t 2

t=1

 n1  c

2

1 2

Now put y = e 

 ne  nx 

x2 2e

2

 ln x 

2



3  xe 3 2

dy   x  1  y  3   dx   x  1   2

10.

 1 

 x  1  3 dy y   dx x  1  x  1



y 3 3  1  x c 2 x 1 x 1  x  1



y   x  c  x  1  3



0   2  c   3  3



c  3



y  x2  2 x

2

 1 

   x 1 dx  e   x 1dx   x  12  3  dx







x 1





VMC/Differential Equation

20

HWT-Solutions/Mathematics

Vidyamandir Classes 2



x

Area in 4th quadrant 

2



 2 x dx 

8 4 3

0

 

4 4  sq. units 3 3

Differential Equation

1.

 y  y 2 cos    xy sin  x   y 2 xy cos    x sin x

dy Now  dx

HWT - 3

y x  y x

Now let y  tx

2.

3.

dt t 2 cos t  t sint  dx t cos t  sint



tx



xdt 2t sint t cos t  sint 2dx   dt   dx t cos t  sint t sint x





Now

y 1 



n



y y  sin  2 x x 2x

 t cos t  sint  dt  t sint



 c  n t sin t  n  2  x 



c

 2



sin

y    k2 x 2 xy

   t 2 2

 1  nc 2

Only B option satisfies this condition. f  x 

x2  1  x 1



f  x 

x x 2  1 1 x2  log  x  x 2  1   c 2 2   2

Since

f  0 

1 2 1 2  c 2 2

Now

f 1 

2

 4.

2dx x



f  x  f  x





2 log 1  2 1 1 2     2 2 2 2 2 1 log 2





2 1



dy y dx



n y  x  c

as

f  0  1



 1

e

dx x

e

x





y   ex



f  x  ex

e x dx

e

2x

1

Let e  k x

 5.(D)

 dy   dy   1  x  y  0    dx   dx 

VMC/Differential Equation

k

dk

 tan 1  k   tan 1 e x

1 dy 1 dx 2

or

dy dy  0 y x

21



y = x + c or xy = c



(x – y + c) (xy – c) = 0

HWT-Solutions/Mathematics

Vidyamandir Classes 6.(C)

x2  y 2  y  tan 1    c 2 x

7.(A)

d ( xy )  x .

8.(C)

dy x 2   dx x 2 

9.(D)



f ( xy ) f ( xy ) dx  d ( xy )  x dx f ( xy ) f ( xy )



y2 y

2

tan 1 y 

 (1  x)dx 



y (0) = 0







log  f ( xy ) 



x2  x  k; 2

x2 c  2



 x2 y  tan  x  2 



 x2   y  tan  x   2  

k = 0 

dy 2 x 2 y  y  ax3  dx x x2  1

10.(D)

 c  x2  y 2   y  xtan    2  



 



2





1.(C)

x x d      . xdx.  y  y

x2  dx ; y

y2

2

x In    x 2  2c  y



2.(BC)

p2 – px + y = 0

2



Put y = tx





4.(A)

dx I .F .  e  x

y

x d   y  x    y



 x  x2 xdx  In    c  y 2

2

2

 e x . C  y 2e x  kx 2

Diff. w.r.t. x to get : - 2 p

dp dp  p x  p0 dx dx

dt 3t (1  t ) (1  t )   dx 2t 2  1

In t



t 2  1  3 n x  const.



y – cx + c2 = 0 is the general solution.



x  x 2 2 x  y  0    

2

x in the original equation, we have 2

x



2

HWT - 4

p = c or x = 2p

Also, substituting p  3.(C)

x2



;

dp  0 or x = 2p  dx





 k  

1  2x  P  x  x  1

1  2 x2 dy ax3  y  dx x x 2  1 x x2  1

Differential Equation ydx  xdy

2 f ( xy )  ke x / 2

dy 1 dx at 1, 0 



2

1  y  dy

 x2  y 2  k   y  x tan    2  

1

1/ 2 





  t  t 1  1  1  dt   x dx  const. 



xy

1/ 2

y

x2 4

3

y 2  x 2  C.

1

5.(C)

dy x y 1  dx 2x  2 y  3 

6.(B)

 enx  x

dt t2  dx 2t  3

Let x + y = t

 1





dy dt  dx dx





1  3 xdx  The solution is : t13e 







2 1  x2 e  x 2  1 e x  C 2 t

7.(C)

xdx 



ydx  xdy y4

0

2



x x x3dx    d    0   y  y



VMC/Differential Equation

(2t  3)dt  dx t2





dt  xt  x3t 3 dx

 1  3 x e 3



dy dt  dx dx

1



Put x + y = t

1  3 xdx

 x  y

x x3dx     y

2

2

2t + loge(t – 2) = x + c 

x + 2y + loge (x + y – 2) = c

[Bernouill’s OE)

dx

1

dt t 1 1  dx 2t  3



;



2 1  x2 e   2 x3e  x dx  C 2 t

2  x 2  1  ce x

ydx  xdy y2

0

3

x4 1  x      c  3 x 4 y 3  4 x3  cy 3 4 3  y

22

HWT-Solutions/Mathematics

Vidyamandir Classes

8.(B)

dy y   dx x

 y f  x  y f   x

Substitute y  vx

dy dv vx dx dx







9. (B)

Equation of tangent to the parabola y2 = 4ax is : y  mx 

a , m is any arbitrary constant and m

 10. (B)

yx

dx

dy  dy  x .   y a0 dx  dx 



HWT - 5

dy e 2 x dy   ex  e y  e y e x  e2 x dx e y dx  dy  dt Put, ey  t  ey     dx  dx

 

dt  t e x  e 2 x where I.F.  e dx



x



 

x

x

e y . ee  ee

x

2

dx  c

 e  1  c x

x 2  y 2  2ay  0 Differentiate w.r.t. x 

2 x  2 py  2ap  0  a 



 e x dx  ee x x

x

x



t . ee  e x . ee  ee  c



e y  e x  1  c . ee

x

. . . .(i)

Substitute the value of a in (i)  2x  x2  y 2    2 y   0   p 

2 x  2 py 2p

x

2



 y 2 p  2 xy



cos y dx  1  2e  x sin y dy  0   

 cos y dy   1  2e  log  cos y   log  e  2   c sin y





dx

x

x

cos y

e

x

At x  0, y 

4.(B)

f (v)

 f (v) dv   x

2

dy a  dx dy dx

t . ee  ee . e x

3.(C)



Differentiate twice



2.(A)

dv f (v )  dx f (v)

dy  m. dx

Differential Equation

1.(A)

x.

 y f    cx x



log f (v) = log x + log c



2



k

 1  k 4 3 2

e x  2  3 2 cos y

dy  yg   x   g  x  . g   x  dx

I.F. = e

 g  x  dx  e g  x 

g x y . e    g  x  . g   x  dx  c



y .e 

g  x

g x g x  g  x e    e    c





g  x   log 1  y  g  x   k

VMC/Differential Equation

23

HWT-Solutions/Mathematics

Vidyamandir Classes 5.(B) 6.(A)

d2y dx 2 x3

8.(AB)

9.(B)

2y  0

dx 2

 xdy  ydx   1  cos y  dx x2





7.(D)

d2y

  2 y 

   y d  x  y 1  cos x

Put x  y  t 

x 

x

1  y   tan    c  2 2 x   2x

dx 3

 x  cex y 1  0

 y xdy  ydx  y log   dx x xdy  ydx y  y  dx   . log   . x x x x2  y d  dx x   x  y  y  x  log  x       y  y Substitute log    t  log    cy x x dy x  y  x y [Linear D.E.] dx 1  x 2

1 dy  x    y x y dx  1  x 2 



y t

Substitute

1 dy dt dt  x    dx  1  x 2 2 y dx dx

 21 x2  dx x

I.F. = e

e

t x   2 2

 

1  ln 1 x 2 4

1

1  x  2

Hence the general solution is : 1 x y .  14 2 1 x 2 1  x2



1  x  2

14









1 1  x2 3



34

 



 x  2 y  dydx  y  y . dydx  x  2 y

3

3





14

dx  c

c

3 y   1  x2  c 1  x2



10.(D)



y





14

14

dx x   2 y2 dt y



 1 y dy

where

I.F. = e



x.

e

log 1 y 



1 y



1 1 x  2 y 2  dy   y2  c y y y

VMC/Differential Equation

24

HWT-Solutions/Mathematics