Jee 2014 Booklet6 Hwt Differential Equations

Vidyamandir Classes DATE :

TIME : 40 Minutes

MARKS : [ ___ /10]

TEST CODE : DE [1]

START TIME :

END TIME :

TIME TAKEN:

PARENT’S SIGNATURE :

 This test contains a total of 10 Objective Type Questions. Each question carries 1 mark. There is NO NEGATIVE marking. Choose the correct alternative. Only one choice is correct. 1.

2.

x

dy  y  x 2  y 2 is the differential equation of the family of curves represented by : dx

(A)

y  x 2  y 2  Cx 2

(B)

x  x 2  y 2  Cy 2

(C)

x  x 2  y 2  Cy

(D)

x  x 2  y 2  Cx

A curve passes through the point (5, 3) and at any point (x, y) on the curve, the product of its slope and the ordinate is equal to its abscissa. The equation of the curve is : (A)

3.

x 1

(B)

x2

x 2  y 2  16

(C)

5 x 2  3 y 2  98

(D)

3x  5 y  0

(D)

x2 x 1

x 2  2 x  1 dy x2  y , is : dx x  x  1 x 1

x 1

(C)

x2

x2 x 1

The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes through the point (4, 3). The equation of the curve is : (A)

5.

(B)

An integrating factor of the equation

(A) 4.

x 2  y 2  34

y2  x  5

(B)

y2  2x  1

(C)

x2  y  5

(D)

2x  5  y

y  C . e2 x  1

(D)

ye x  2 x  C

 e 2 x y  dx   1 is : General solution of   x x  dy   (A)

y  2 x e 2 x  C (B)

y  x  e 2 x  C (C) 2

6.

dy  dy  If the curves given by the solutions of x     y  x   y  0 are passing through (2, 3) then another point of intersection is: dx  dx 

(A) 7.

8.

 3, 2 

(B)

 3,  2 

The general solution of the differential equation

(C)

(3, 2)

 3,  2 

dy  y tan x  y 2 sec x is : dx

(A)

tan x   c  sec x  y

(B)

sec y   c  tan y  x

(C)

sec x   c  tan x  y

(D)

None of these

If

(D)

dy  3 x 2 y 2  3 x 2  y 2  1 ; y  0   0 . Then y equals : dx

  1  tan x .tan  x  tan  x  x  tan x  tan x3

(A)

(C)

3

1

3

VMC/Differential Equations

(B)

(D)

116

tan x  tan x3 1  tan x .tan x3

tan 1 x  tan 1 x3

HWT-6/Mathematics

Vidyamandir Classes 9.

The differential equation of the family of curves y  Ae3 x  Be5 x , where A, B are arbitrary constants is : (A)

d2y dx

2

2

(C)

10.

d y dx 2

8 5

dy  15 y  0 dx

2

dx (D)

d y dx 2

1 dy 1 ecos x is :  y dx 1  x2 1  x2

2 e 1 x

e 2

VMC/Differential Equations

(B)

d2y 2

dy  6y  0 dx

An integrating factor of

(A)

(B)

 2

e

.e

 cos 1 x

2

8

8

dy  5y  0 dx

dy  15 y  0 dx

 

sin 1 x 2

(C)

117

e

e 2

(D)

None of these

HWT-6/Mathematics

Vidyamandir Classes DATE :

TIME : 40 Minutes

MARKS : [ ___ /10]

TEST CODE : DE [2]

START TIME :

END TIME :

TIME TAKEN:

PARENT’S SIGNATURE :

 This test contains a total of 10 Objective Type Questions. Each question carries 1 mark. There is NO NEGATIVE marking. Choose the correct alternative. Only one choice is correct. 1.

The differential equation whose general solution is y 2  x  1  my where m is an arbitrary constant, is : (A)

2.

3.

(A)



(C)



dx

2

x

x2

dy  y  0 (C) dx

d2y

x2

x

dy y0 dx

dx

2



2 x3  2   log x   c 3  3 

y2 x2 y2



2 x3  2   log x   c 3  3 

(B)



x3  2   log x   c 3  3 

(D)

None of these



 dydx

y  x  y2  1

(D)

(C)

ey

y2

 2 x

x2

(B)

 y2

   3x



x2 y2

2

(D)

ex

2

 2 y

   3x

2 y 2  x2  3 y

(B) (D)

If for the differential equation y  

(A)

y2

x2  y 2 and it passes through the point (2, 1). The equation of the curve is : 2 xy

2 x2  y 2  3 y 2

x2

dx 2 x   10 y 2 , is : dy y

The slope of a curve at any point (x, y) on it is

(C)

6.

d2y

(B)

An integrating factor of the equation

(A)

5.

dy 2 dx

Solution of the equation xdy   y  xy 3 1  log x   dx  0 is :  

(A)

4.

yx

2

 x2

x x y x     the general solution is y  then    is given by : log Cx x  y  y

(B)

y2 x2

x2

(C)

y2



(D)



y2 x2



The equation of the curve satisfying the differential equation y2 x 2  1  2 xy1 passing through the point (0, 1) and having slope of tangent at x = 0 as 3 is : (A)

7.

(B)

General solution of the equation x cos (A)

8.

y  x 2  3x  2

xy  k sec xy

(B)

y  x3  3 x  1

(C)

y  x 2  3x  1

y y  ydx  xdy   y sin  xdy  ydx  x x y x xy  k sec (C) xy  k sec x y

(D)

None of these

(D)

xy  k cosec

y x

The differential equation of the family of curves represented by y 3  cx  c3  c 2  1 , where c is an arbitrary constant is of (A)

order 1, degree 1

VMC/Differential Equations

(B)

order 2, degree 1

(C)

118

order 1, degree 2

(D)

order 2, degree 2

HWT-6/Mathematics

Vidyamandir Classes 9.

2e  1

(A)

10.





The solution of the initial value problem x 2  y 2 dy  xy dx, y 1  1 is y  y  x  . If y  x0   e , then x0 equals : (B)

e 3

(C)

e2  1

(D)

A curve passes through (2, 0) and the slope of the tangent at P(x, y) on the curve is given by

e 3

 x  12  y  3 . Then, the area  x  1

bounded by the curve and the x-axis in the 4th quadrant is : (A)

2 3

VMC/Differential Equations

(B)

4 3

(C)

119

1

(D)

5 3

HWT-6/Mathematics

Vidyamandir Classes DATE :

TIME : 40 Minutes

MARKS : [ ___ /10]

TEST CODE : DE [3]

START TIME :

END TIME :

TIME TAKEN:

PARENT’S SIGNATURE :

 This test contains a total of 10 Objective Type Questions. Each question carries 1 mark. There is NO NEGATIVE marking. Choose the correct alternative. Only one choice is correct. 1.

 y y   If the differential equation y  cos   x dy  y dx   x  sin   x dy  y dx   0 with y 1  has the solution in the form 2 x x    

 y  sin , then k equals : k xy x (A) 2.

2

y  C1 x5  C2

If f   x   (A)

4.

1

(C)

3

(D)

1 2

Equation of the curve y  f  x  , satisfying xy5  y4 (suffixes denoting differentiation) & which is symmetric with respect to yaxis, is: (A)

3.

(B)

1 2

x 1  x 1 log 2  1 2





log e 2 x  1  c

(C)

y  C1 x 4  C2

(D)

y  C1 x 4  C2 x 2  C3

2 1

(D)

 log

 

(D)

None of these

1 2 then f (1) equals : 2

and f  0  

1 log 2

(B)

If f   x   f  x  , f  0   1 , then (A)

y  C1 x 2  C2

(B)

2 1

(C)

log

 f  x  f x 

2 1

dx

log (e 2 x  e  x )  c (C)

(B)

tan 1 e x  c

2

5.

dy  dy  Solution of the equation x     y  x   y  0 is : dx  dx  (A) (B)  x  2 y  c  xy  c   0 (C)

6.

(C)

If x (A)

8.

(D)

Solution of the equation xdx  ydy 

(A)

7.

 x  y  c  2 xy  c   0 x2  y 2

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

 0 is :

(B)

 x  x2  y 2 x  y tan   2 

(D)

None of these

   

f  x .y dy yx. then f  x . y  is equal to : (k being an arbitrary constant) dx f  x . y  ke x

2 2

Integral curve satisfying y   (A)

xdy  ydx

 x  y  c  xy  c   0  x  y  c  xy  c   0



5 3

VMC/Differential Equations

ke y

(B)

x y 2

2

x2  y 2 (B)

2 2

(C)

ke xy

2

(D)

None of these

, y 1  1 has the slope at the point (1, 0) of the curve equal to :

1

(C)

120

1

(D)

5 3

HWT-6/Mathematics

Vidyamandir Classes 9.

10.

The solution of y   1  x  y 2  xy 2 , y  0   0 is : (A)

 x2  y 2  exp . x   1  2  

(C)

y  tan c  x  x 2



(B)



(D)



 



The integrating factor of x 1  x 2 dy  2 x 2 y  y  ax3 dx  0 is e (A)

2 x2  a2



x 1  x2



VMC/Differential Equations

(B)

2 x3  1

(C)

121

 x2  y 2  1  c exp . x    2    x2  y 2  tan  x    2  

 Pdx , then P is : 2 x2  1 ax3

(D)

2 x2  1



x 1  x2



HWT-6/Mathematics

Vidyamandir Classes DATE :

TIME : 40 Minutes

MARKS : [ ___ /10]

TEST CODE : DE [4]

START TIME :

END TIME :

TIME TAKEN:

PARENT’S SIGNATURE :

 This test contains a total of 10 Objective Type Questions. Each question carries 1 mark. There is NO NEGATIVE marking. Choose the correct alternative. Only one choice is correct. 1.

Solution of y dx  x dy  x 2 y dx is : (A)

2

ye x  cx 2

(B)

2

ye  x  cx 2

2

2

y 2 e x  cx 2

(D)

y 2 e  x  cx 2

(C)

y  2x  4

(D)

y  2 x2  4

(C)

xy y 2  x 2  C

(D)

x y

(C)

x

(D)

ex

(C)

2

2.

 dy   dy  A solution of the differential equation :    x    y  0 dx    dx  (A)

3.

y

3

y2

(B)

 

4 y  x2



 2 x 2 y dx  2 xy 2  x3 dy  0 represent the curve :

x 2 x  y 2  C (B) xy x 2  y 2  C y dy y   x3  3 ;  x  0  is : Integrating factor of dx x (A)

4.

(A) 5.

6.

y  x  log  x  y  2   c

(B)

y  2 x  log  x  y  2   c

(C)

2 y  x  log  x  y  2   c

(D)

2 y  2 x  log  x  y  2   c

The solution of the

1  x 2  1  Ce x x y 1

 x  y

2

(B)

 x  1  Ce x



(D)

1

 x  y

2

 x 2  1  Ce x

2

2 1  x  1  Ce x x y

4 x y  3 x  cy 4 3

3



3

(B)

3 x3 y 4  4 x3  cy 3 (C)

3 x 4 y 3  4 x3  cy 3 (D)

None of these

 y f    cxy x

(D)

None of these

(D)

None of these

 f  y x  Solution of the equation xdy   y  x  dx is :    y x   f 

(A)

10.

dy 3  x  x  y   x3  x  y   1 is : dx

Solution of equation xy 4  y dx  xdy  0 is : (A)

9.

log x

(A)

(C)

8.

(B)

Solution of  x  y  1 dx   2 x  2 y  3 dy  0 is :

(A)

7.

x

y 2  x2  C

x f    cy  y

(B)

 y f    cx x

(C)

The degree of the differential equation of all tangent lines to the parabola y 2  4ax is : (A) 1 (B) 2 (C) 3

The order and degree of the differential equation, of which xy  ce x  be  x  x 2 is a solution, is : (A) 1, 3 (B) 2, 1 (C) 3, 2 (D)

VMC/Differential Equations

122

None of these

HWT-6/Mathematics

Vidyamandir Classes DATE :

TIME : 40 Minutes

MARKS : [ ___ /10]

TEST CODE : DE [5]

START TIME :

END TIME :

TIME TAKEN:

PARENT’S SIGNATURE :

 This test contains a total of 10 Objective Type Questions. Each question carries 1 mark. There is NO NEGATIVE marking. Choose the correct alternative. Only one choice is correct. 1.

2.

Solution of the equation (A)

e y  e x  1  ce e

x

(C)

e x  e y  1  ce e

y

(C)

5.

ex – 2 = 3 2 cos y ex + 2 = 3 2 cos y

g  x   log 1  y  g  x    C

(C)

g  x   log 1  y  g  x    C

(D)

None of these

x

(x2 + y2) y = 2xy

 is : 4

(B)

ex + 2 =

(D)

None of these

2 cos y

dy  yg   x   g  x  . g   x  where g(x) is a given function of x, is : dx g  x   log 1  y  g  x    C (B)

(D)

None of these

Differential equation for y  Acos  x  B sin  , x , where A and B are the arbitrary constants, is : d2y dx 2

 2 y  0

(B)

d2y dx 2

2 y  0

dy y  xy  1  cos is : dx x 1  y  tan    c  2  2x  2x c  y cos    1  x x

(C)

d2y dx 2

 y  0

(D)

d2y dx 2

 y  0

The solution of x 2

(C)

The solution of (A)

8.

e x  e y  1  ce e

(D)

(A)

(A)

7.

2 (x2 – y2) y = xy

The general solution of the differential equation

(A)

6.

(B)

The particular solution of cos y dx + 1  2e  x  sin y dy = 0 when x = 0, y = (A)

4.



The differential equation for the family of curves x2 + y2 – 2ay = 0, where a is an arbitrary constant is : (A) (x2 – y2) y = 2 xy (B) 2 (x2 + y2) y = xy (C)

3.



dy  e x  y e x  e y is : dx

y 1 c x x

(B)

tan

(D)

x 2  c  x 2 tan

(C)

 x  cex y  0

(D)

 x  c ex y 1  0

x  cx y

(D)

log





y x

dy  1  e x  y is : dx

 x  y  ex y  0

(B)

 x  cex y  0

dy  y  log y  log x  1 , then the solution of the equation is : dx y x log  cy (A) (B) (C) log  cy x y

If x

VMC/Differential Equations

123

log

y  cx x

HWT-6/Mathematics

Vidyamandir Classes 9.

10.

Solution of differential equation

dy x  y  x y is : dx 1  x 2

    y  1  x   c 1  x 

(A)

3 y  1  x2  c 1  x2

(C)

3

2

14

2



 dydx  y is : x  y c  y 



 

(B)

3 y   1  x2  c 1  x2

(D)

None of these

(C)

x  2 y c  y2



14

Solution of the differential equation x  2 y 3 (A)



x  y2 c  y2



VMC/Differential Equations

(B)

2

124





(D)



x  y c  y2



HWT-6/Mathematics