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

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

Solution:

we have dy/dx = 2y + z and ...(i)

dz/dx = 3y ...(ii)

from 3 (i) l (ii), we get

implies

on integrating, we get

implies ln(3y + z) = 3x + c

So, 3y + z = ke^{3x}

QUESTION: 2

One of the integrating factors of the differential equation (y^{2 }- 3xy)dx + (x^{2} - xy)dy = 0 is

Solution:

Since, given differential equation is homogeneous

So, I.F =

QUESTION: 3

The differential equation 2ydx - (3y - 2x)dy = 0

Solution:

Here

QUESTION: 4

The solution of the differential equation

Solution:

we have

QUESTION: 5

Consider the differential equation 2 cos (y^{2})dx - xy sin (y^{2})dy = 0

Solution:

we have 2cos (y^{2})dx - xy sin (y^{2}) dy = 0 Multiplying by x^{3}

and

Since,

So, x^{3} is integrating factor.

QUESTION: 6

Let f, g : [-1, 1] → R, f (x) = x^{3}, g(x) = x^{2} IxI Then

y" + py' + qy = 0 on [-1, 1]

Solution:

Here, lxI = -x for x < 0 = x for x ≥ 0

So, g(x) = x^{3} for x ≥ 0

= -x^{3} for x < 0

and f(x) = x^{3}

QUESTION: 7

The solution of the differential equation C

Solution:

The given differential equation is homogeneous

Multiplying the differential equations by I.F.

...(i)

...(ii)

Adding (i) and (ii) writing common terms once

implies

QUESTION: 8

An integration factor of is

Solution:

we have

QUESTION: 9

The general solution of

(Here c_{1} and c_{2} are arbitrary constants.)

Solution:

Let z = log x, Then

The differential equation changes to

implies

So,

QUESTION: 10

If y = (c_{1} + c_{2}x ) e^{x} is the general solution of the differential equation then k equals

Solution:

Let log x = z, Then

The differential equation changes to

QUESTION: 11

Let y_{1}(x) and y_{2}(x) be linearly independent solution of the differential equation y" + P(x) y' + Q(x)y = 0, where P(x) and Q(x) are continuous functions on an interval I. Then y_{3}(x) = ay_{1}(x) + by_{2}(x) and y_{4}(x) = cy_{1}(x) dy_{2(}x) are linearly independent solutions of the given differential equations if

Solution:

y_{3}(x) and y_{4}(x) are linearly dependent if y_{3}(x) = ky_{4}(x) where k is any constant

Since, y_{1}(x) and y_{2}(x) are independent

So, a - kc = 0

and b - kd = 0

or

So, y_{3}(x) and y_{4}(x) are independent if

implies ad ≠ bc

QUESTION: 12

Consider the differential equations, satisfying y(0) = 0, where This initial value problem

Solution:

we have

or,

Thus,

implies y = x^{2} + c for x ≤ 0,

but y(0) = 0 implies c = 0

So,

QUESTION: 13

The particular integral of the differential equation

y" + y' + 3y = 5 cos(2x + 3) is

Solution:

we have (D^{2} + D + 3)y = 5 cos(2x + 3)

So,

QUESTION: 14

The general solution of differential equation

4x^{2}y" - 8xy' + 9y = 0 is

Solution:

Let z = log x, Then

So, The differential equation becomes

So,

QUESTION: 15

Let 1, x and x^{2} be the solutions of a second order linear non-homogeneous differential equation on -1 < x < 1. Then its general solution, involving arbitrary constants c_{1} and c_{2}, can be written as

Solution:

QUESTION: 16

If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x^{2}, then the general solution of the differential equation is

Solution:

g(x, y)dx + (x + y) dy = 0 is exact if

implies

g (x , 0) = x^{2}

So, g(x, y) = x^{2} + y

So, general solution is

implies 2x^{3} + 6xy + 3y^{2} = c

QUESTION: 17

A general solution of the differential equation

Solution:

QUESTION: 18

Consider the differential equation Then is equal to

Solution:

we have

implies

QUESTION: 19

Consider the differential equation dy/dx = ay - by^{2}, where a, b > 0 and y(0) = y_{0}. As x → + ∝, the solution y (x) tends to

Solution:

So, Equation becomes,

QUESTION: 20

Orthogonal trajectories of the family of curves (x - 1)^{2} + y^{2} + 2ax = 0 are the solution of the differential equation

Solution:

we have (x - 1)^{2} + y^{2} + 2ax = 0 ...(i)

Differentiating w.r.t. x, we get

2(x - 1) + 2y y' + 2a = 0

implies

Putting expression of a in eq. (i), we get

(x - 1)^{2} + y^{2} - 2x((x - 1) + yy') = 0

implies x^{2} - 2x + 1 + y^{2} - 2x^{2} + 2x + 2xyy' = 0

Orthogonal trajectories are given by

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