Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.7 | Set 1

Last Updated : 05 Mar, 2021

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) = (e^x+ 1)y

Solution:

We have,
(dy/dx) = (e^x+ 1)y
(dy/y) = (e^x+ 1)dx
On integrating both sides
∫(dy/y) = ∫(e^x+ 1)dx
log(y) = (e^x+ 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) = e^xy⁴

Solution:

We have,
5(dy/dx) = e^xy⁴
5(dy/y⁴) = e^x
On integrating both sides,
5∫(dy/y⁴) = ∫e^x
-(5/3)(1/y³) = e^x+ c (Where ‘c’ is integration constant)

Question 7. xcosydy = (xe^xlogx + e^x)dx

Solution:

We have,
xcosydy = (xe^xlogx + e^x)dx
cosydy = e^x(logx + 1/x)dx
On integrating both sides,
∫cosydy = ∫e^x(logx + 1/x)dx
Since, ∫[f(x) + f'(x)]e^xdx] = e^xf(x)
siny = e^xlogx + c (Where ‘c’ is integration constant)

Question 8. (dy/dx) = e^x+y+ x²e^y

Solution:

We have,
(dy/dx) = e^x+y+ x²e^y
(dy/dx) = e^xe^y+ x²e^y
dy = e^y(e^x+ x²)dx
e^-ydy = (e^x+ x²)dx
On integrating both sides,
∫e^-ydy = ∫(e^x+ x²)dx
-e^-y= e^x+ (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. (e^y+ 1)cosxdx + e^ysinxdy = 0

Solution:

We have,
(e^y+ 1)cosxdx + e^ysinxdy = 0
(cosx/sinx)dx = -[e^y/(e^y+ 1)]dy
On integrating both sides,
∫(cosx/sinx)dx = -∫[e^y/(e^y+ 1)]dy
log(sinx) = -log(e^y+ 1) + log(c)
log(sinx) + log(e^y+ 1) = log(c)
log[sinx(e^y+ 1)] = log(c)
sinx(e^y+ 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) = (xe^xlogx + e^x)/(xcosy)

Solution:

We have,
(dy/dx) = (xe^xlogx + e^x)/(xcosy)
xcosydy = (xe^xlogx + e^x)dx
cosydy = e^x(logx + 1/x)dx
On integrating both sides,
∫cosydy = ∫e^x(logx + 1/x)dx
Since, ∫[f(x) + f'(x)]e^xdx] = e^xf(x)
siny = e^xlogx + c (Where ‘c’ is integration constant)

Question 15. (dy/dx) = e^x+y+ x³e^y

Solution:

We have,
(dy/dx) = e^x+y+ x³e^y
(dy/dx) = e^xe^y+ x³e^y
dy = e^y(e^x+ x³)dx
e^-ydy = (e^x+ x³)dx
On integrating both sides,
∫e^-ydy = ∫(e^x+ x³)dx
-e^-y= e^x+ (x⁴/4) + c
e^-y+ e^x+ (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 = e^x(sin²x + sin2x)dx
On integrating both sides,
∫y(2logy + 1)dy = ∫e^x(sin²x + sin2x)dx
$2logy*∫ydy-∫[2\frac{d}{dx}(logy)∫ydy]dy+\frac{y^2}{2}=e^xsin^2x+c$
Since, ∫e^x(sin²x + sin2x)dx = e^xsin²x
Using property ∫[f(x) + f'(x)]e^x = e^xf(x)
y²log(y) – ∫ydy + y²/2 = e^xsin²x + c
y²log(y) – y²/2 + y²/2 = e^xsin²x + c
y²log(y) = e^xsin²x + 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)

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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.7 | Set 1

Solve the following differential equations:

Question 1. (x – 1)(dy/dx) = 2xy

Question 2. (x2 + 1)dy = xydx

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

Question 4. (x – 1)(dy/dx) = 2x3y

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

Question 6. 5(dy/dx) = exy4

Question 7. xcosydy = (xexlogx + ex)dx

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

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

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

Question 11. xcos2ydx = ycos2xdy

Question 12. xydy = (y – 1)(x + 1)dx

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

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

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

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

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

Question 18.

Question 19.

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

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Question 2. (x²+ 1)dy = xydx

Question 3. (dy/dx) = (e^x+ 1)y

Question 4. (x – 1)(dy/dx) = 2x³y

Question 5. xy(y + 1)dy = (x²+ 1)dx

Question 6. 5(dy/dx) = e^xy⁴

Question 7. xcosydy = (xe^xlogx + e^x)dx

Question 8. (dy/dx) = e^x+y+ x²e^y

Question 9. x(dy/dx) + y = y²

Question 10. (e^y+ 1)cosxdx + e^ysinxdy = 0

Question 11. xcos²ydx = ycos²xdy

Question 14. (dy/dx) = (xe^xlogx + e^x)/(xcosy)

Question 15. (dy/dx) = e^x+y+ x³e^y

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

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

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

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