RD Sharma Solutions Class 12 Differential Equations Exercise 22.9

Download PDF of RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 9

 

rd sharma solutions for class 12 maths chapter 22 ex 9 1
rd sharma solutions for class 12 maths chapter 22 ex 9 2
rd sharma solutions for class 12 maths chapter 22 ex 9 3
rd sharma solutions for class 12 maths chapter 22 ex 9 4

 

Access RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 9

EXERCISE 22.9

Question. 1

Solution:

From the question it is given that, x2 dy + y (x + y) dx = 0

The given differential equation can be written in standard form as,

dy/dx = – (y(x + y))/x2

So, it is a homogeneous equation,

Let us assume, y = vx and (dy/dx) = v + x (dv/dx)

Then, v + x (dv/dx) = – (vx (x + vx))/x2

v + x (dv/dx) = – v – v2

Transposing,

x (dv/dx) = – v – v – v2

x (dv/dx) = – 2v – v2

Now, taking like variables on same side,

dv/(2v + v2) = – dx/x

Integrating on both side we get,

∫(1/(2v + v2)dv = – ∫dx/x

∫(1/(v2 + 2v + 1 – 1)dv = – ∫dx/x

We know that, (a + b)2 = a2 + 2ab + b2

∫(1/((v + 1)2 – 12)dv = – ∫dx/x

½ log [(v + 1 – 1)/(v + 1 + 1)] = – log x + log c

Log [v/(v + 2)]1/2 = – log (c/x)

v/(v + 2) = c2/x2

((y/x)/((y/x) + 2)) = c2/x2

By simplification we get,

y/(y + 2x) = c2/x2

yx2 = (y + 2x) c2

Question. 2

Solution:

From the question it is given that, (dy/dx) = (y – x)/(y + x)

The given differential equation is a homogeneous equation,

Let us assume, y = vx and (dy/dx) = v + x (dv/dx)

v + x (dv/dx) = (vx – x)/(vx + x)

Then, v + x (dv/dx) = (v – 1)/(v + 1)

x (dv/dx) = ((v – 1)/(v + 1)) – v

x (dv/dx) = (v – 1 – v2 – v)/(v + 1)

On dividing we get,

x (dv/dx) = – (1 + v2)/(v + 1)

Now, taking like variables on same side,

((v + 1)/(v2 + 1)) = – dx/x

Integrating on both side we get,

∫((v + 1)/(v2 + 1)) dv = – ∫dx/x

∫(v/(v2 + 1)) dv + ∫(1/(v2 + 1)) dv = – ∫dx/x

½ ∫(2v/(v2 + 1)) dv + ∫(1/(v2 + 1)) dv = – ∫dx/x

½ log [v2 + 1] + tan-1 v = – log x + log c

Then, log [(y2 + x2)/x2] + 2 tan-1 (y/x) = 2 log (c/x)

log [x2 + y2] – 2 log x + 2 tan-1 (y/x) = 2 log (c/x)

log (x2 + y2) + 2 tan-1 (y/x) = 2 log c

log (x2 + y2) + 2 tan-1 (y/x) = k

Question. 3

Solution:

From the question it is given that, (dy/dx) = (y2 – x2)/(2xy)

The given differential equation is a homogeneous equation,

Let us assume, y = vx and (dy/dx) = v + x (dv/dx)

v + x (dv/dx) = (v2x2 – x2)/(2xvx)

Then, x (dv/dx) = ((v2 – 1)/(2v)) – (v/1)

x (dv/dx) = (v2 – 1 – 2v2)/(2v)

x(dv/dx) = (-1 – v2)/2v

Now, taking like variables on same side,

((2v)/(v2 + 1)) dv = – dx/x

Integrating on both side we get,

∫((2v)/(v2 + 1)) dv = – ∫dx/x

log (1 + v2) = – log x + log c

1 + v2 = c/x

Now substitute the value of v,

1 + y2/x2 = c/x

x2 + y2 = cx

Question. 4

Solution:

From the question it is given that, x (dy/dx) = (x + y)/x

The given differential equation can be written in standard form as,

dy/dx = (x + y)/x

So, it is a homogeneous equation,

Let us assume, y = vx and (dy/dx) = v + x (dv/dx)

dy/dx = v + x (dv/dx)

Then, v + x (dv/dx) = (x + vx)/x

v + x (dv/dx) = (x/x) + (vx/x)

v + x (dv/dx) = 1 + v

Now, taking like variables on same side,

dv = dx/x

Integrating on both side we get,

∫dv = ∫dx/x

v = log x + c

Now substitute the value of v,

y/x = log x + c

y = x log x + cx

Question. 5

Solution:

From the question it is given that, (x2 – y2)dx – 2xy dy = 0

The given differential equation can be written in standard form as,

dy/dx = (x2 – y2)/2xy

So, it is a homogeneous equation,

Let us assume, y = vx and (dy/dx) = v + x (dv/dx)

v + x (dv/dx) = (x2 – v2x2)/(2xvx)

Then, x (dv/dx) = ((1 – v2)/(2v)) – (v/1)

x (dv/dx) = (1 – v2 – 2v2)/(2v)

x(dv/dx) = (1 – 3v2)/2v

Now, taking like variables on same side,

((2v)/(1 – 3v2)) dv = dx/x

Integrating on both side we get,

∫((2v)/(1 – 3v2)) dv = – ∫dx/x

1/-3∫((-6v)/(1 – 3v2)) dv = – 3∫dx/x

log (1 – 3v2) = -3 log x + log c

1 – 3v2 = c/x3

Now substitute the value of v,

x3 (1 – (3y2/x2)) = c/

x3 (x2 – 3y2)/x2 = c

x(x2 – 3y2) = c

Leave a Comment

Your email address will not be published. Required fields are marked *