Class 12 RD Sharma Solutions Chapter 22 Differential Equations Exercise 22.9 | Set 1 (original) (raw)

Last Updated : 28 Aug, 2024

Solve the following differential equations

Question 1. x2dy + y(x + y)dy = 0

**Solution:

We have,

x2dy + y(x + y)dy = 0

dy/dx = -y(x + y)/x2

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = -vx(x + vx)/x2

v + x(dv/dx) = -v - v2

x(dv/dx) = -2v - v2

\frac{dv}{(v^2 + 2v)} = -\frac{dx}{x}

On integrating both sides,

∫\frac{dv}{(v^2+2v+1-1)}=-∫(\frac{dx}{x})

∫\frac{dv}{(v+1)^2-(1)^2}=-log(x)

\frac{1}{2}log(\frac{v+1-1}{v+1+1})=-log|x|+log|c|

log|v/(v + 2)|1/2 = -log|x/c|

v/(v + 2) = c2/x2

\frac{\frac{y}{x}}{\frac{y}{x}+2}=\frac{c^2}{x^2}

yx2 = (y + 2x)c2 (Where ‘c’ is integration constant)

Question 2. (dy/dx) = (y - x)/(y + x)

**Solution:

We have,

(dy/dx) = (y - x)/(y + x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

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

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)

x(dv/dx) = -(v2 + 1)/(v + 1)

\frac{(v+1)dv}{(v^2+1)}=-\frac{dx}{x}

On integrating both sides,

∫\frac{(v+1)dv}{(v^2+1)}=-∫\frac{dx}{x}

∫vdv/(v2+1)+∫dv/(v2+1)=-∫(dx/x)

\frac{1}{2}∫\frac{2v}{v^2+1}+∫\frac{dv}{v^2+1}=-log|x|+log|c|

(1/2)log|v2 + 1| + tan-1(v) = log(c/x)

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

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

log(y2 + x2) + 2tan-1(y/x) = 2log(c) (Where ‘c’ is integration constant)

Question 3. (dy/dx) = (y2 - x2)/2yx

**Solution:

We have,

(dy/dx) = (y2 - x2)/2yx

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = (v2x2 - x2)/2vx2

v + x(dv/dx) = (v2 - 1)/2v

x(dv/dx) = [(v2 - 1)/2v] - v

x(dv/dx) = (v2 - 1 - 2v2)/2v

x(dv/dx) = -(v2 + 1)/2v

\frac{2vdv}{(v^2 + 1)} = -\frac{dx}{x}

On integrating both sides,

∫\frac{vdv}{(v^2+1)}=-∫\frac{dx}{x}

log|v2+1| = -log(x) + log(c)

log|v2+1| = log(c/x)

y2/x2 + 1 = |c/x|

(x2 + y2) = cx (Where ‘c’ is integration constant)

Question 4. x(dy/dx) = (x + y)

**Solution:

We have,

x(dy/dx) = (x+y)

(dy/dx) = (x+y)/x

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

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

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

x(dv/dx) = 1

dv = (dx/x)

On integrating both sides,

∫dv = ∫(dx/x)

v = log(x) + c

y/x = log(x) + c

y = xlog(x) + cx (Where ‘c’ is integration constant)

Question 5. (x2 - y2)dx - 2xydy = 0

**Solution:

We have,

(x2 - y2)dx - 2xydy = 0

(dy/dx) = (x2 - y2)/2xy

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = (x2 - v2x2)/2xvx

v + x(dv/dx) = (1 - v2)/2v

x(dv/dx) = [(1 - v2)/2v] - v

x(dv/dx) = (1 - 3v2)/2v

\frac{2vdv}{(1-3v^2)} = \frac{dx}{x}

On integrating both sides,

∫\frac{2vdv}{(1-3v^2)}=∫\frac{dx}{x}

\frac{1}{3}∫\frac{6v}{1-3v^2}dv=∫\frac{dx}{x}

-(1/3)log(1 - 3v2) = log(x) - log(c)

log(1 - 3v2) = -log(x)3 + log(c)

(1-\frac{3y^2}{x^2})=(\frac{c}{x^3})

(x2 - 3y2)/x2 = (c/x3)

x(x2 - 3y2) = c (Where ‘c’ is integration constant)

Question 6. (dy/dx) = (x + y)/(x - y)

**Solution:

We have,

(dy/dx) = (x + y)/(x - y)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

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

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

x(dv/dx) = [(1 + v)/(1 - v)] - v

x(dv/dx) = (1 + v - v + v2)/(1 - v)

x(dv/dx) = (1 + v2)/(1 - v)

\frac{(1-v)dv}{(v^2+1)}=\frac{dx}{x}

On integrating both sides,

∫\frac{(v+1)dv}{(v^2+1)}=∫\frac{dx}{x}

∫dv/(v2 + 1) - ∫vdv/(v2 + 1) = ∫(dx/x)

tan-1(v) - (1/2)log(v2 + 1) = log(x) + c

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

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

tan-1(y/x) = (1/2)log(y2 + x2) + c (Where ‘c’ is integration constant)

Question 7. 2xy(dy/dx) = (x2 + y2)

**Solution:

We have,

2xy(dy/dx) = (x2 + y2)

(dy/dx) = (x2 + y2)/2xy

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = (x2 + v2x2)/2xvx

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

x(dv/dx) = [(1 + v2)/2v] - v

x(dv/dx) = (1 - v2)/2v

\frac{2vdv}{(1 - v^2)} = \frac{dx}{x}

On integrating both sides,

∫\frac{2vdv}{(1 - v^2)} = ∫\frac{dx}{x}

-log(1 - v2) = log(x) - log(c)

log(1 - v2) = -log(x) + log(c)

1 - y2/x2 = (c/x)

(x2 - y2) = cx (Where ‘c’ is integration constant)

Question 8. x2(dy/dx) = x2 - 2y2 + xy

**Solution:

We have,

x2(dy/dx) = x2 - 2y2 + xy

(dy/dx) = (x2 - 2y2 + xy)/x2

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = (x2 - 2v2x2 + xvx)/2xvx

v + x(dv/dx) = (1 - 2v2 + v)/x2

x(dv/dx) = (1 - 2v2 + v) - v

x(dv/dx) = (1 - 2v2)

dv/(1 - 2v2) = (dx/x)

On integrating both sides,

dv/(1 - 2v2) = ∫(dx/x)

∫\frac{dv}{v^2-\frac{1}{2}}=-2\frac{dx}{x}

∫\frac{dv}{(\frac{1}{\sqrt2})^2-v^2}=2\frac{dx}{x}

\frac{1}{\sqrt{2}}log|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}|=2log(x)+log(c)

\frac{1}{\sqrt{2}}log|\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}|=2log(x)+log(c)

(\frac{1}{√2})log|\frac{(x + y√2)}{(x - y√2)}|=log(x)^2 + log(c)

|\frac{(x + y√2)}{(x - y√2)}|^{\frac{1}{√2}} = cx^2

|\frac{(x+y√2)}{(x-y√2)}|=|cx^2|^{√2} (Where ‘c’ is integration constant)

Question 9. xy(dy/dx) = x2 - y2

**Solution:

We have,

xy(dy/dx) = x2 - y2

(dy/dx) = (x2 - y2)/xy

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = (x2 - v2x2)/xvx

v + x(dv/dx) = (1 - v2)/v

x(dv/dx) = [(1 - v2)/v] - v

x(dv/dx) = (1 - 2v2)/v

vdv/(1 - 2v2) = (dx/x)

On integrating both sides,

∫vdv/(1 - 2v2) = ∫(dx/x)

∫4vdv/(1 - 2v2) = 4∫(dx/x)

-log(1 - 2v2) = 4log(x) - log(c)

log(1 - 2v2) = log(c/x4)

(1 - 2y2/x2) = c/x4

(x2-2y2)/x2 = c/x4

x2(x2 - 2y2) = c (Where ‘c’ is integration constant)

Question 10. yex/ydx = (xex/y + y)dy

**Solution:

We have,

yex/ydx = (xex/y + y)dy

(dy/dx) = (xex/y + y)/yex/y

It is a homogeneous equation,

So, put x = vy (i)

On differentiating both sides w.r.t x,

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

So,

v + y(dv/dy) = (vyevy/y + y)/yevy/y

v + y(dv/dy) = (vev + 1)/ev

y(dv/dy) = [(vev + 1)/ev] - v

y(dv/dy) = (vev + 1 - vev)/ev

y(dv/dy) = (1/ev)

evdv = (dy/y)

On integrating both sides,

∫evdv = ∫(dy/y)

ev = log(y) + log(c)

ex/y = log(y) + log(c) (Where ‘c’ is integration constant)

Question 11. x2(dy/dx) = x2 + xy + y2

**Solution:

We have,

x2(dy/dx) = x2 + xy + y2

dy/dx = (x2 + xy + y2)/x2

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = (x2 + xvx + v2x2)/x2

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

x(dv/dx) = (1 + v + v2) - v

dv/(1 + v2) = (dx/x)

On integrating both sides,

∫dv/(1 + v2) = ∫(dx/x)

tan-1(v) = log|x| + c

tan-1(y/x) = log|x| + c (Where ‘c’ is integration constant)

Question 12. (y2 - 2xy)dx = (x2 - 2xy)dy

**Solution:

We have,

(y2 - 2xy)dx = (x2 - 2xy)dy

(dy/dx) = (y2 - 2xy)/(x2 - 2xy)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = (v2x2 - 2xvx)/(x2 - 2xvx)

v + x(dv/dx) = (v2 - 2v)/(1 - 2v)

x(dv/dx) = [(v2 - 2v - v + 2v2)/(1 - 2v)]

x(dv/dx) = 3(v2 - 1)/(1 - 2v)

-(2v - 1)dv/(v2 - v) = 3(dx/x)

On integrating both sides,

-∫(2v - 1)dv/(v2 - v) = 3∫(dx/x)

-log|v2 - v| = 3log|x| - log|c|

log|v2 - v| = log|c/x3|

(y2/x2 - y/x) = (c/x3)

(y2 - xy) = c/x

x(y2 - xy) = c (Where ‘c’ is integration constant)

Question 13. 2xydx + (x2 + 2y2)dy = 0

**Solution:

We have,

2xydx + (x2 + 2y2)dy = 0

dy/dx = -(2xy)/(x2 + 2y2)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,

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

So,

v + x(dv/dx) = -(2xvx)/(x2 + 2v2x2)

v + x(dv/dx) = -(2v)/(1 + 2v2)

x(dv/dx) = -[(2v)/(1 + 2v2)] - v

x\frac{dv}{dx}=\frac{-3v-2v^3}{1+2v^2}

\frac{1+2v^2}{3v+2v^3}dv=-\frac{dx}{x}

On integrating both sides,

∫\frac{1+2v^2}{3v+2v^3}dv=-∫\frac{dx}{x}

Substituting (3v + 2v3) = z

On differentiating both sides w.r.t x,

3(1 + 2v)dv = dz

(1 + 2v)dv = (dz/3)

(1/3)∫(dz/z) = -∫(dx/x)

(1/3)log|z| = -log|x| + log|c|

log|3v + 2v3| = log|c/x|3

3y/x + 2(y/x)3 = (c/x)3

(3yx2 + 2y3) = c (Where ‘c’ is integration constant)

Summary

Exercise 22.9 | Set 1 typically focuses on solving first-order linear differential equations. These equations are of the form:

dy/dx + P(x)y = Q(x)

where P(x) and Q(x) are functions of x, and y is a function of x.

**Key points:

1. The general solution is found using the integrating factor method.

2. The integrating factor is e^(∫P(x)dx).

3. The general solution formula is: y = e^(-∫P(x)dx) [∫Q(x)e^(∫P(x)dx)dx + C]

4. This method transforms the equation into an exact differential equation.

Practice Questions

1. Solve: dy/dx + 3y = e^(2x)

2. Find the general solution of: dy/dx - 2y = x^2

3. Solve the initial value problem: dy/dx + y = 2x, y(0) = 1

4. Find a particular solution of: x dy/dx + y = x^3, given that y = 2 when x = 1

5. Solve: dy/dx + (1/x)y = x^2, x > 0

6. Find the general solution of: dy/dx + y tan(x) = sec(x)

7. Solve: (1 - x^2) dy/dx - xy = x^3

8. Find the particular solution of: dy/dx - y = e^x, given that y = 0 when x = 0

9. Solve the differential equation: dy/dx + 2xy = x

10. Find the general solution of: dy/dx + y/x = ln(x), x > 0