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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

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, e^{z} = 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 e^{z})

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 (y^{2})dx - xy sin (y^{2})dy = 0

Solution:

**Correct Answer :- D**

**Explanation : **2cos (y^{2})dx - xy sin (y^{2}) dy = 0

Multiplying by x^{3}

2x^{3} cos(y^{2})dx - x^{4}y sin(y^{2})dy = 0

dM/dy = - 4x^{3}y sin(y^{2})

dN/dx = - 4x^{3}y sin(y^{2})

Since dM/dy = dN/dx

So, x^{3} is an 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

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

Solution:

Let z = log x, Then

The differential equation changes to

implies

So,

QUESTION: 10

Solution of the differential equation

Solution:

=> x^{2} lnx dy - xy dy=xy dx – y^{2 }lny dx …….dividing by x^{2} y^{2} then

on integrating we get

where c is a constant of integration.

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

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 2^{nd} order.

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

x^{2} - y^{2} - 1 - 2xydy/dx = 0(d)

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