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This mock test of Differential Equations - 2 for IIT JAM helps you for every IIT JAM entrance exam.
This contains 20 Multiple Choice Questions for IIT JAM Differential Equations - 2 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The solution of the differential equation

Solution:

we have (D^{3} - 9D)y = cos x

implies D(D^{2} - 9)y = cos x

So, complete solution is

QUESTION: 2

Consider the differential equation y" + 6y' + 25y - 0 with initial condition y(0) = 0. Then, the general solution of the IVP is

Solution:

we have (D^{2} + 6D + 25)y = 0

So, y =e^{-3x} [A cos 4x + B sin 4x]

but y(0) = 0 implies A = 0

Hence, y = Be^{-3x} sin 4x

QUESTION: 3

The solution of the differential equation y" + 4y = 0 subject to y(0) = 1, y' (0) = 2 is

Solution:

we have (D^{2} + 4) = 0

So, solutions is y = c_{1} sin 2x + c_{2} cos 2x

but y(0)= 1 implies c_{2} = 1

and y'(0) = 2 implies 2c_{1}

= 2 implies c_{1} = 1

Hence y = sin 2x + cos 2x

QUESTION: 4

General solution of the differential equation is given by

Solution:

we have xdy = (y + xe^{-y/x})dx

implies xdy - ydx = xe^{-y/x} dx

QUESTION: 5

The Wronskian of the function x |x| is zero for

Solution:

QUESTION: 6

The general solution of y" - m^{2}y = 0 is

Solution:

**Correct Answer :- A**

**Explanation **: y''- m^{2}y = 0

r^{2} - m^{2} = 0

r = +-m

y = Ae^{m}x + Be^{-m}x

Let A=C+D and B=D-C

= Ce^{m}x + De^{m}x - Ce^{-m}x + De^{-m}x

= C(e^{m}x-e^{-m}x) + D(e^{m}x+e^{-m}x)

= Csinh(mx) + Dcosh(mx)

QUESTION: 7

The orthogonal trajectories of the curves y^{2} = 3x^{3}+x + c are

Solution:

Differentiating w.r.t. x, we get

Replacing the equation of orthogonal trajectory is given by

implies

Integrating both sides, we get

implies 2 tan^{-1} 3x + 3 In |y| = k

QUESTION: 8

The general solution of the differential equation

Solution:

is satisfied, So, equation is exacct

...(i)

...(ii)

Adding (i) and (ii), writing the common term once.

QUESTION: 9

The solution of the initial value problem xy' - y = 0 with y(1) = 1 is

Solution:

QUESTION: 10

If y'_{1} (x) = 3y_{1}(x) + 4y_{2}(x) and y'_{2} (x) = 4y_{1}(x) + 3y_{2}(x), then y_{1}(x) is

Solution:

we have (D - 3)y_{1} - 4y_{2} = 0

4y_{2} + (3 - D)y_{1} = 0

Solving equations, we get

So,

QUESTION: 11

If e^{x} + xy + x sin y + e^{y} = c is the general solution of an exact differential equation, then the differential equation is

Solution:

Taking total differential

QUESTION: 12

The solution of the differential equation

Solution:

we have

and

Since,

So, differential equation is exact

Now,

and

So, solution of differential equation

QUESTION: 13

The differential equation (2x^{2} + by^{2})dx + cxydy = 0 is made exact by multiplying the integrating factor 1/x^{2}. Then the relation between and c is

Solution:

Multiplying the differential equation by 1/x^{2}, we get

It is exact

So,

implies

2b + c = 0

QUESTION: 14

If general solution of the differential equation ay'" + by" + cy’ + dy = 0 is linearly spanned by e^{x}, sinx and cosx, then which one of the following holds?

Solution:

QUESTION: 15

If y = In (sin (x + a)) + b, where a and b are constants, is the primitive, then the corresponding lowest order differential equation is

Solution:

we have y = In sin (x + a ) + b

implies y'= cot (x + a)

implies y" = - cosec^{2} (x + a)

QUESTION: 16

The solution of the initial value problem

Solution:

QUESTION: 17

What is the degree of the non-homogeneous partial differential equation,

Solution:

Degree of an equation is defined as the power of the highest derivative present in the equation. Hence from the equation, the degree is 5.

QUESTION: 18

Two linearly independent solutions of the differential equation y" - 2y' + y = 0 are y_{1} = e^{x} and y_{2} = xe^{x}. Then a particular solution of y" - 2y' + y = e^{x} sin x is

Solution:

QUESTION: 19

Consider the differential equation ( x + y + 1) dx + (2x + 2y + 1) dy = 0. Which of the following statements is true?

Solution:

QUESTION: 20

Which one of the following differential equations represents all circles with radius a?

Solution:

(x - h)^{2} + (y - k)^{2} = a^{2} ...(i)

where h, k are parameters

Differentiating w.r.t. x, we get

( x - h ) + ( y - k ) y' = 0 ...(ii)

Differentiating w.r.t. x again 1 + ( y - k)

y^{11} + y^{12 } = 0

implies (iii)

From (ii), we get ( x - h ) = - ( y - k)y'

...(iv)

Putting (x - h) and (y - k) in (i), we get

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