Class 12 RD Sharma Solutions Chapter 22 Differential Equations Exercise 22.7 | Set 1 (original) (raw)

Last Updated : 30 Aug, 2024

Solve the following differential equations:

Question 1. (x - 1)(dy/dx) = 2xy

**Solution:

We have,

(x - 1)(dy/dx) = 2xy

dy/y = [2x/(x - 1)]dx

On integrating both sides,

∫(dy/y) = ∫[2x + (x - 1)]dx

log(y) = ∫[2 + 2/(x - 1)]dx

log(y) = 2x + 2log(x - 1) + c (Where 'c' is integration constant)

Question 2. (x2 + 1)dy = xydx

**Solution:

We have,

(x2 + 1)dy = xydx

(dy/y) = [x/(x2 + 1)]dx

On integrating both sides

∫(dy/y) = ∫[x/(x2 + 1)]dx

log(y) = (1/2)∫[2x/(x2 + 1)]dx

log(y) = (1/2)log(x2 + 1) + c (Where 'c' is integration constant)

Question 3. (dy/dx) = (ex + 1)y

**Solution:

We have,

(dy/dx) = (ex + 1)y

(dy/y) = (ex + 1)dx

On integrating both sides

∫(dy/y) = ∫(ex + 1)dx

log(y) = (ex + x) + c (Where 'c' is integration constant)

Question 4. (x - 1)(dy/dx) = 2x3y

**Solution:

We have,

(x - 1)(dy/dx) = 2x3y

(dy/y) = [2x3/(x - 1)]dx

On integrating both sides

∫(dy/y) = ∫[2x3/(x - 1)]dx

∫(dy/y) = 2∫[x2 + x + 1 + 1/(x - 1)]dx

log(y) = (2/3)(x3) + x2 + 2x + 2log(x - 1) + c (Where 'c' is integration constant)

Question 5. xy(y + 1)dy = (x2 + 1)dx

**Solution:

We have,

xy(y + 1)dy = (x2 + 1)dx

y(y + 1)dy = [(x2 + 1)/x]dx

(y2 + y)dy = xdx + (dx/x)

On integrating both sides,

∫(y2 + y)dy = ∫xdx + (dx/x)

(y3/3) + (y2/2) = (x2/2) + log(x) + c (Where 'c' is integration constant)

Question 6. 5(dy/dx) = exy4

**Solution:

We have,

5(dy/dx) = exy4

5(dy/y4) = ex

On integrating both sides,

5∫(dy/y4) = ∫ex

-(5/3)(1/y3) = ex + c (Where 'c' is integration constant)

Question 7. xcosydy = (xexlogx + ex)dx

**Solution:

We have,

xcosydy = (xexlogx + ex)dx

cosydy = ex(logx + 1/x)dx

On integrating both sides,

∫cosydy = ∫ex(logx + 1/x)dx

Since, ∫[f(x) + f'(x)]exdx] = exf(x)

siny = exlogx + c (Where 'c' is integration constant)

Question 8. (dy/dx) = ex+y + x2ey

**Solution:

We have,

(dy/dx) = ex+y + x2ey

(dy/dx) = exey + x2ey

dy = ey(ex + x2)dx

e-ydy = (ex + x2)dx

On integrating both sides,

∫e-ydy = ∫(ex + x2)dx

-e-y = ex + (x3/3) + c (Where 'c' is integration constant)

Question 9. x(dy/dx) + y = y2

**Solution:

We have,

x(dy/dx) + y = y2

x(dy/dx) = y2 - y

[1/(y2 - y)]dy = dx/x

On integrating both sides,

∫[1/(y2 - y)]dy = ∫dx/x

∫[1/(y - 1) - 1/y]dy = ∫(dx/x)

log(y-1) - log(y) = logx + logc

log[(y - 1)/y] = log[xc]

(y - 1)/y = xc

(y-1) = yxc (Where 'c' is integration constant)

Question 10. (ey + 1)cosxdx + eysinxdy = 0

**Solution:

We have,

(ey + 1)cosxdx + eysinxdy = 0

(cosx/sinx)dx = -[ey/(ey + 1)]dy

On integrating both sides,

∫(cosx/sinx)dx = -∫[ey/(ey + 1)]dy

log(sinx) = -log(ey + 1) + log(c)

log(sinx) + log(ey + 1) = log(c)

log[sinx(ey + 1)] = log(c)

sinx(ey + 1) = c (Where 'c' is integration constant)

Question 11. xcos2ydx = ycos2xdy

**Solution:

We have,

xcos2ydx = ycos2xdy

(x/cos2x)dx = (y/cos2y)dy

xsec2xdx = ysec2ydy

On integrating both sides,

∫xsec2xdx = ∫ysec2ydy

x∫sec^2x - ∫[\frac{d(x)}{dx(x)}∫sec^2xdx]dx = y∫sec^2x-∫[\frac{d(y)}{dy}(y)∫sec^2y]dy

xtanx - ∫tanxdx = ytany - ∫tanydy

xtanx - log(secx) = ytany - log(secy) + c (Where 'c' is integration constant)

Question 12. xydy = (y - 1)(x + 1)dx

**Solution:

We have,

xydy = (y - 1)(x + 1)dx

[y/(y - 1)]dy = [(x + 1)/x]dx

On integrating both sides,

∫[y/(y - 1)]dy = ∫[(x + 1)/x]dx

∫[1 + 1/(y - 1)]dy = ∫[(x + 1)/x]dx

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

y - x = log(x) - log(y - 1) + c (Where 'c' is integration constant)

Question 13. x(dy/dx) + coty = 0

**Solution:

We have,

x(dy/dx) + coty = 0

x(dy/dx) = -coty

dy/coty = -(dx/x)

tanydy = -(dx/x)

On integrating both sides,

∫tanydy = -∫(dx/x)

log(secy) = -log(x) + log(c)

log(secy) + log(x) = log(c)

log(xsecy) = log(c)

x/cosy = c

x = c * cosy (Where 'c' is integration constant)

Question 14. (dy/dx) = (xexlogx + ex)/(xcosy)

**Solution:

We have,

(dy/dx) = (xexlogx + ex)/(xcosy)

xcosydy = (xexlogx + ex)dx

cosydy = ex(logx + 1/x)dx

On integrating both sides,

∫cosydy = ∫ex(logx + 1/x)dx

Since, ∫[f(x) + f'(x)]exdx] = exf(x)

siny = exlogx + c (Where 'c' is integration constant)

Question 15. (dy/dx) = ex+y + x3ey

**Solution:

We have,

(dy/dx) = ex+y + x3ey

(dy/dx) = exey + x3ey

dy = ey(ex + x3)dx

e-ydy = (ex + x3)dx

On integrating both sides,

∫e-ydy = ∫(ex + x3)dx

-e-y = ex + (x4/4) + c

e-y + ex + (x4/4) = c (Where 'c' is integration constant)

Question 16. y√(1 + x2) + x√(1 + y2)(dy/dx) = 0

**Solution:

We have,

y√(1 + x2) + √(1 + y2)(dy/dx) = 0

y√(1 + x2)dx = -x√(1 + y2)dy

\frac{\sqrt{1+y^2}}{y}dy=-\frac{\sqrt{1+x^2}}{x}dx

\frac{y\sqrt{1+y^2}}{y^2}dy=-\frac{x\sqrt{1+x^2}}{x^2}dx

On integrating both sides,

∫\frac{y\sqrt{1+y^2}}{y^2}dy=-∫\frac{x\sqrt{1+x^2}}{x^2}dx

Let, 1 + y2 = z2

On differentiating both sides

2ydy = 2zdz

ydy = zdz

= ∫\frac{y\sqrt{1+y^2}}{y^2}dy

= ∫\frac{zdz*z}{z^2-1}

= ∫[z2/(z2 - 1)]dz

= ∫[1 + 1/(z2 - 1)]dz

= z + (1/2)log[(z - 1)/(z + 1)]

On putting the value of z in above equation

= \sqrt{1+y^2}+\frac{1}{2}log[\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}]

Similarly,

∫\frac{x\sqrt{1+x^2}}{x^2}dx = \sqrt{1+x^2}+\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]

\sqrt{1+y^2}+\frac{1}{2}log[\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}]=-\sqrt{1+x^2}-\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]+c

\sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}log[\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}]+\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]=c (Where 'c' is integration constant)

Question 17. √(1 + x2)(dy) + √(1 + y2)dx = 0

**Solution:

We have,

√(1 + x2)(dy) + √(1 + y2)dx = 0

\frac{dy}{\sqrt{(1+y^2)}}=\frac{-dx}{\sqrt{(1+x^2)}}

On integrating both sides,
log[y + √(1 + y2)] = -log[x + √(1 + x2)] + logc

log[y + √(1 + y2)] + log[x + √(1 + x2)] = logc

log([y + √(1 + y2)][x + √(1 + x2)]) = logc

[y + √(1 + y2)][x + √(1 + x2)] = c (Where 'c' is integration constant)

Question 18. \sqrt{(1 + x^2 + y^2 + x^2y^2)} + xy(\frac{dy}{dx}) = 0

**Solution:

We have,

\sqrt{(1 + x^2 + y^2 + x^2y^2)} + xy(\frac{dy}{dx}) = 0

\sqrt{[(1 + x^2) + y^2(1+x^2)]}+xy(\frac{dy}{dx})=0

\sqrt{[(1+x^2)(1+y^2)]}+xy(\frac{dy}{dx})=0

\frac{y}{\sqrt{1+y^2}}dy=-\frac{\sqrt{1+x^2}}{x}dx

\frac{y}{\sqrt{1+y^2}}dy=-\frac{x\sqrt{1+x^2}}{x^2}dx

On integrating both sides,

∫\frac{y}{\sqrt{1+y^2}}dy=-∫\frac{x\sqrt{1+x^2}}{x^2}dx

Let, 1 + x2 = z2

On differentiating both sides

2xdx = 2zdz

xdx = zdz

= -∫\frac{x\sqrt{1+x^2}}{x^2}dy

= -∫\frac{zdz*z}{z^2-1}

= -∫[z2/(z2 - 1)]dz

= -∫[1 + 1/(z2 - 1)]dz

= -z - (1/2)log[(z - 1)/(z + 1)]

On putting the value of z in above equation

=-\sqrt{1+x^2}-\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]

Let, 1 + y2 = v2

On differentiating both sides

2ydy = 2vdv

ydy = vdv

= ∫(vdv/v)

= v

On putting the value of v in above equation

= √(1 + y2)

= \sqrt{1+y^2}=-\sqrt{1+x^2}-\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]+c

= \sqrt{1+y^2}+\sqrt{1+x^2}+\frac{1}{2}log[\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}]=c (Where 'c' is integration constant)

Question 19. (\frac{dy}{dx}) = \frac{e^x(sin^2x + sin2x)}{y(2logy+1)}

**Solution:

We have,

(\frac{dy}{dx}) = \frac{e^x(sin^2x + sin2x)}{y(2logy+1)}

y(2logy + 1)dy = ex(sin2x + sin2x)dx

On integrating both sides,

∫y(2logy + 1)dy = ∫ex(sin2x + sin2x)dx

2logy*∫ydy-∫[2\frac{d}{dx}(logy)∫ydy]dy+\frac{y^2}{2}=e^xsin^2x+c

Since, ∫ex(sin2x + sin2x)dx = exsin2x

Using property ∫[f(x) + f'(x)]ex = exf(x)

y2log(y) - ∫ydy + y2/2 = exsin2x + c

y2log(y) - y2/2 + y2/2 = exsin2x + c

y2log(y) = exsin2x + c (Where 'c' is integration constant)

Question 20. (dy/dx) = x(2logx + 1)/(siny + ycosy)

**Solution:

We have,

(dy/dx) = x(2logx + 1)/(siny + ycosy)

(siny + ycosy)dy = x(2logx + 1)dx

On integrating both sides,

∫(siny + ycosy)dy = ∫x(2logx + 1)dx

∫sinydy + y∫cosydy - ∫{(dy/dy)∫cosydy}dy = 2logx∫xdx - 2∫{\frac{d(logx)}{dx}∫xdx} + ∫xdx

-cosy + ysiny - ∫sinydy = x2logx - ∫xdx + (x2/2) + c

-cosy + ysiny + cosy = x2logx - (x2/2) + (x2/2) + c

ysiny = x2logx + c (Where 'c' is integration constant)

Summary

Exercise 22.7, Set 1 focuses on solving homogeneous differential equations of the first order. These equations are characterized by the right-hand side being expressible as a function of y/x. The problems range from simple rational expressions to more complex forms involving quadratic terms. Students are asked to find both general and particular solutions. The primary method used to solve these equations involves the substitution y = vx, where v is a new function of x, which transforms the homogeneous equation into a separable one. This exercise helps students recognize homogeneous forms, apply the appropriate substitution, and solve the resulting separable equations.