# Differential Equations - 1

## 20 Questions MCQ Test IIT JAM Mathematics | Differential Equations - 1

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Attempt Differential Equations - 1 | 20 questions in 60 minutes | Mock test for IIT JAM preparation | Free important questions MCQ to study IIT JAM Mathematics for IIT JAM Exam | Download free PDF with solutions
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 = ke3x

QUESTION: 2

### Find the General Solution of the given differential equation. Solution:

This special case where D has a co-efficient is solved using Legendre’s method.

Let log⁡(8x + 7) = z

Then, ez = 8x + 7 Substituting this in the equation,

8Dy + 2y = x

y(8D + 2) = x

Thus the auxiliary equation is 8D + 2 = 0

Thus, D = -1 / 4   (Solving by substituting powers of ez) The General solution is C.F + P.I = 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 (y2)dx - xy sin (y2)dy = 0

Solution:

Explanation : 2cos (y2)dx - xy sin (y2) dy = 0

Multiplying by x3

2x3 cos(y2)dx - x4y sin(y2)dy = 0

dM/dy = - 4x3y sin(y2)

dN/dx = - 4x3y sin(y2)

Since dM/dy = dN/dx

So, x3 is an integrating factor.

QUESTION: 6

Let f, g : [-1, 1] → R, f (x) = x3, g(x) = x2 IxI Then
y" + py' + qy = 0 on [-1, 1]

Solution:

Here, lxI = -x for x < 0 = x for x ≥ 0
So, g(x) = x3 for x ≥ 0
= -x3 for x < 0
and f(x) = x3

QUESTION: 7

If the general solutions of a differential equation are (y + c)2 = cx, where c is an arbitrary constant, then the order and degree of differential equation is

Solution:

There will only one constant in the first-order differential equation. Differentiating the given equation. Putting the value of c in Eq. (1) and simplifying we will get a first-order and second-degree equation. Hence, (A) is the correct answer.

QUESTION: 8

An integration factor of is

Solution:

we have QUESTION: 9

The general solution of (Here c1 and c2 are arbitrary constants.)

Solution:

Let z = log x, Then The differential equation changes to implies So, QUESTION: 10

Solution of the differential equation Solution: => x2 ln⁡x dy - xy dy=xy dx – y2 ln⁡y dx …….dividing by x2 y2 then  on integrating we get  where c is a constant of integration.

QUESTION: 11

Let y1(x) and y2(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 y3(x) = ay1(x) + by2(x) and y4(x) = cy1(x) dy2(x) are linearly independent solutions of the given differential equations if

Solution:

y3(x) and y4(x) are linearly dependent if y3(x) = ky4(x) where k is any constant Since, y1(x) and y2(x) are independent
So, a - kc = 0
and b - kd = 0
or So, y3(x) and y4(x) are independent if QUESTION: 12

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

Solution:

we have or, Thus, implies y = x2 + 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 (D2 + D + 3)y = 5 cos(2x + 3)
So,  QUESTION: 14

The general solution of differential equation
4x2y" - 8xy' + 9y = 0 is

Solution:

Let z = log x, Then  So, The differential equation becomes So, QUESTION: 15

What is the order of the non-homogeneous partial differential equation, Solution:

The order of an equation is defined as the highest order derivative present in the equation and hence from the equation, it is clear that it of 2nd order.

QUESTION: 16

If g(x, y)dx + (x + y )dy = 0 is an exact differential equation and if g(x, 0) = x2, 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) = x2
So, g(x, y) = x2 + y So, general solution is implies 2x3 + 6xy + 3y2 = 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 - by2,  where a, b > 0 and y(0) = y0.  As  x → + ∝, the solution y (x) tends to

Solution: So, Equation becomes,   QUESTION: 20

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

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