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# Differential Equations - 12

## 20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Differential Equations - 12

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This mock test of Differential Equations - 12 for Mathematics helps you for every Mathematics entrance exam. This contains 20 Multiple Choice Questions for Mathematics Differential Equations - 12 (mcq) to study with solutions a complete question bank. The solved questions answers in this Differential Equations - 12 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this Differential Equations - 12 exercise for a better result in the exam. You can find other Differential Equations - 12 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.
QUESTION: 1

Solution:

QUESTION: 2

### Let k be a real constant. The solution of the differential equations  satisfies the relation

Solution:

Differentiating equation (i) w.r.t. we get

QUESTION: 3

### If general solution of the differential equation ay",+ by"- cy’+ dy = 0 is linearly spanned by ex sin x and cos x, then which one of the following holds?

Solution:

Linearly independent solutions of ay" + by”+cy' + dy = 0 are ex cos x and sin x, so roots of aD3 + bD2 + cD + d = 0 will be 1 and ±f, so we have 1 +

QUESTION: 4

Two linearly independent solutions of the differential equation y" - 2y' + y = 0 are y1 = ex and y2 = xex. Then a particular solution of y2 - 2y' + y + ex sin x is

Solution:

Particular integral of y" - 2y' + y - ex sin x is

QUESTION: 5

Orthogonal trajectories of the family of curves (x - 1)2 + y2 + 2ax = 0 are the solutions of the differential equation

Solution:

By putting the value of a from equation (ii) in equation (i), we get.

It ’s orthogonal trajectory will be D.E.

QUESTION: 6

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

Solution:

Equation of the circles with radius ‘a' is

Putting values of (y - k) and (x - h) from equation (iv) and equation (v) in equation (i), we get

QUESTION: 7

The solution of the differential equation  with initial condition y(0) = 0 is

Solution:

QUESTION: 8

The differential equation (2a2 + by2) dx + cxydy = 0 is made exact by multiplying the integrating factor   Then the relation between b and c is.

Solution:

Integrating factor of (2x2 + by2) dx + cxy dy = 0 is

QUESTION: 9

Given that   Which one of the following is always true?

Solution:

....(i)
Differentiating both sides w.r.t. x, we get.

QUESTION: 10

The sum of the intercepts made on the axes of co-ordinates by any tangent to the curve √x + √y = 2 is equal to

Solution:

Equation of tangent is (Y - y) = y'(X - x)

Intercept on Y-axis.

Intercept on X-axis:

QUESTION: 11

Solution of the equation

Solution:

0

QUESTION: 12

If  and it is known that for a = 1, y = 1; if x = -1, then the value of y will be :

Solution:

QUESTION: 13

The solution of the differential equation

Solution:

QUESTION: 14

If the solution of the differential equation  is x + y -1 = Ceu then the value of u is :

Solution:

Correct Answer :- C

Explanation : dy/dx = (x+y-z)/(x+y)

Let t = x + y

1 + dy/dx = dt/dx

Equation becomes,

dt/dx - 1 = (t-z)/t

dt/dx = (2t-z)/t

tdt/(2t-z) = dx

= 1/2[1+z/(2t-z)]dt = dx

t + 1/2log(2t-z) = 2x+c

2t-z = e2(x-y+c)

x+y-1 = cex-y

QUESTION: 15

The solution of the equation

Solution:

which is linear differential equation so, it’s solution is

QUESTION: 16

The solution of the differential equation

Solution:

which is linear differential equation in dependent variable x and independent variable z.
whose solution is

QUESTION: 17

The solution of the equation

Solution:

.......(i)
which is linear differential equation, so solution is

QUESTION: 18

The degree and order of differential equation  are respectively

Solution:

QUESTION: 19

The solution of the equation (2x + y + 1)dx + (4x + 2y – 1)dy = 0 is:

Solution:

QUESTION: 20

The solution of the differential equation y(x2y + ex)dx- exdy = 0 is

Solution:

Correct Answer :- D

Explanation : y(x2y + ex)dx - exdy = 0

Dividing it by y2

(x2 + ex/y)dx - ex/y2 dy = 0

x2 dx + (yex - ex dy)/y2 = 0

d(x3/3) + d(ex/y) = 0

∫d(x3/3) + ∫d(ex/y) = 0

=> (x3/3) + (ex/y) = c

=> x3y + 3ex = 3yc