RD Sharma Solutions Class 12 Differential Equations Exercise 22.9

RD Sharma Solutions for Class 12 Maths Exercise 22.9 Chapter 22 Differential Equations are available here. While solving the exercise questions from the RD Sharma book, students often face difficulty and eventually pile up their doubts. Our expert tutors formulate this RD Sharma Solutions for Class 12 Maths Chapter 22 to assist you with your exam preparation and help you attain good marks in Maths.

RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 9

Download PDF

carouselExampleControls112

Previous Next

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 the same side,

dv/(2v + v²) = – dx/x

Integrating on both sides 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 the same side,

((v + 1)/(v² + 1)) = – dx/x

Integrating on both sides 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 the same side,

((2v)/(v² + 1)) dv = – dx/x

Integrating on both sides 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 the same side,

dv = dx/x

Integrating on both sides 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 the same side,

((2v)/(1 – 3v²)) dv = dx/x

Integrating on both sides 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