RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	27
Exercise	27.7
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations

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)

Solve the following differential equations:

Question 21. (1 – x²)dy + xydx = xy²dx

Solution:

We have,

(1 – x²)dy + xydx = xy²dx

(1 – x²)dy = xy²dx – xydx

(1 – x²)dy = xy(y – 1)dx

$\frac{dy}{y(y-1)}=\frac{xdx}{(1-x^2)}$

On integrating both sides,

$∫[\frac{1}{y-1}-\frac{1}{y}]dy=\frac{1}{2}∫\frac{2x}{1-x^2}dx$

log(y – 1) – logy = -(1/2)log(1 – x²) + logc

log(y – 1) – logy + (1/2)log(1 – x²) = logc (Where ‘c’ is integration constant)

Question 22. tanydx + sec²ytanxdy = 0

Solution:

We have,
tanydx + sec²ytanxdy = 0
tanydx = -sec²ytanxdy
(sec²y/tany)dy = -dx/tanx
On integrating both sides,
∫(sec2y/tany)dy = -∫cotxdx
Let, tany = z
On differentiating both sides
sec²xdx = dz
∫(dz/z) = -∫cotxdx
log(z) = -log(sinx) + log(c)
On putting the value of z in above equation
log(tany) + log(sinx) = log(c)
log[(sinx)(tany)] = log(c)
sinx.tany = c (Where ‘c’ is integration constant)

Question 23. (1 + x)(1 + y²)dx + (1 + y)(1 + x²)dy = 0

Solution:

We have,

(1 + x)(1 + y²)dx + (1 + y)(1 +x²)dy = 0

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

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

On integrating both sides,

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

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

Question 24. tany(dy/dx) = sin(x + y) + sin(x – y)

Solution:

We have,
tany(dy/dx) = sin(x + y) + sin(x – y)
tany(dy/dx) = 2sin{(x + y + x – y)/2}cos{(x + y – x + y)/2}
tany(dy/dx) = 2sinxcosy
(tany/cosy)dy = 2sinxdx
On integrating both sides,
∫secytanydy = 2∫sinxdx
secy = -2cosx + c
secy + cosx = c (Where ‘c’ is integration constant)

Question 25. cosxcosy(dy/dx) = -sinxsiny

Solution:

We have,
cosxcosy(dy/dx) = -sinxsiny
(cosy/siny)dy = -(sinx/cosx)dx
cotydy = -tanxdx
On integrating both sides,
∫cotydy = -∫tanxdx
log(siny) = log(cosx) + logc
log(siny) = log(cosx.c)
siny = c.cosx (Where ‘c’ is integration constant)

Question 26. (dy/dx) + cosxsiny/cosy = 0

Solution:

We have,
(dy/dx) + cosxsiny/cosy = 0
(dy/dx) = -cosx.tany
dy/tany = -cosxdx
cotydy = -cosxdx
On integrating both sides,
∫cotydy = -∫cosxdx
log(cosy) = -sinx + c
log(cosy) + sinx = c (Where ‘c’ is integration constant)

Question 27. x√(1 – y²)(dx) + y√(1 – x²)dy = 0

Solution:

We have,

x√(1 – y²)(dx) + y√(1 – x²)dy = 0

x√(1 – y²)(dx) = -y√(1 – x²)dy

$\frac{ydy}{\sqrt{(1-y^2)}} = \frac{-xdx}{\sqrt{(1 - x^2)}}$

On integrating both sides,

$\frac{1}{2}∫\frac{2ydy}{\sqrt{(1-y^2)}}=-\frac{1}{2}∫\frac{xdx}{\sqrt{(1-x^2)}}$

√(1 – y²) = -√(1 – x²) + c

√(1 – y²) + √(1 – x²) = c (Where ‘c’ is integration constant)

Question 28. y(1 + e^x)dy =(y + 1)e^xdx

Solution:

We have,

y(1 + e^x)dy =(y + 1)e^xdx

$\frac{ydy}{(y+1)}=\frac{e^xdx}{(1+e^x)}$

$1-\frac{1}{(y+1)}=\frac{e^xdx}{(1+e^x)}$

On integrating both sides,

∫[1 – 1/(y + 1)]dy = ∫e^xdx/(1 + e^x)

y – log(y + 1) = log(1 + e^x) + c (Where ‘c’ is integration constant)

Question 29. (y + xy)dx + (x – xy²)dy = 0

Solution:

We have,
(y + xy)dx + (x – xy²)dy = 0
y(1 + x)dx = -x(1 – y²)dy
[(1 – y²)/y]dy = -[(1 + x)/x]dx
On integrating both sides,
∫[(1 – y²)/y]dy = -∫[(1 + x)/x]dx
∫(dy/y) – ∫ydy = -∫dx/x – ∫dx
log(y) – (y²/2) = -log(x) – x + c
log(x) + x + log(y) – (y²/2) = c (Where ‘c’ is integration constant)

Question 30. (dy/dx) = 1 – x + y – xy

Solution:

We have,
(dy/dx) = 1 – x + y – xy
(dy/dx) = (1 – x) + y(1 – x)
(dy/dx) = (1 – x)(1 – y)
dy/(1 – y) = (1 – x)dx
On integrating both sides,
∫dy/(1 – y) = ∫(1 – x)dx
log(1 – y) = x – (x²/2) + c (Where ‘c’ is integration constant)

Question 31. (y²+ 1)dx – (x²+ 1)dy = 0

Solution:

We have,

(y²+ 1)dx – (x²+ 1)dy = 0

(y²+ 1)dx = (x²+ 1)dy

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

On integrating both sides,

$∫\frac{dy}{(y^2+1)}=\frac{∫dx}{(x^2+1)}$

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

Question 32. dy + (x + 1)(y + 1)dx = 0

Solution:

We have,
dy + (x + 1)(y + 1)dx = 0
dy/(y + 1) = -(x + 1)dx
On integrating both sides,
∫dy/(y + 1) = -∫(x + 1)dx
log(y + 1) = -(x²/2) – x + c
log(y + 1) + (x²/2) + x = c (Where ‘c’ is integration constant)

Question 33. (dy/dx) = (1 + x²)(1 + y²)

Solution:

We have,

(dy/dx) = (1 + x²)(1 + y²)

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

On integrating both sides,

$∫\frac{dy}{(1+y^2)}=∫(1+x^2)dx$

tan^-1y = x + (x³/3) + c

tan^-1y – x – (x³/3) = c (Where ‘c’ is integration constant)

Question 34. (x – 1)(dy/dx) = 2x³y

Solution:

We have,

(x – 1)(dy/dx) = 2x³y

dy/y = 2x³dx/(x – 1)

On integrating both sides,

∫dy/y = 2∫x³dx/(x – 1)

$log(y)=2∫[x^2+x+1+\frac{1}{(x-1)}]$

log(y) = (2/3)(x³) + 2(x²/2) + 2x + 2log(x – 1) + log(c)

$log(y)=loge^{[\frac{2}{3}x^3+x^2+2x]}+log(x-1)^2+logc$

y = c|x – 1|²e^{[(2/3)x3+x2+2x]}(Where ‘c’ is integration constant)

Question 35. (dy/dx) = e^x+y+ e^-x+y

Solution:

We have,
(dy/dx) = e^x+y+ e^-x+y
(dy/dx) = e^x.e^y+ e^-x.e^y
(dy/dx) = e^y(e^x+ e^-x)
dy/e^y= (e^x+ e^-x)dx
On integrating both sides,
∫e^-ydy = ∫e^xdx + ∫e^-xdx
-e^-y= e^x– e^-x+ c
e^-x-e^-y= e^x+ c (Where ‘c’ is integration constant)

Question 36. (dy/dx) = (cos²x – sin²x)cos²y

Solution:

We have,
(dy/dx) = (cos²x – sin²x)cos²y
dy/cos²y = (cos²x – sin²x)dx
sex²ydy = cos2xdx
On integrating both sides,
∫sex²ydy = ∫cos2xdx
tany = (sin2x/2) + c (Where ‘c’ is integration constant)

Question 37(i). (xy²+ 2x)(dx) + (x²y + 2y)dy = 0

Solution:

We have,

(xy²+ 2x)(dx) + (x²y + 2y)dy = 0

x(y²+ 2)(dx) = -y(x²+ 2)dy

$\frac{ydy}{(y^2+1)}=\frac{-xdx}{(x^2+1)}$

Multiplying both sides by 2,

$\frac{2ydy}{(y^2+1)}=\frac{-2xdx}{(x^2+1)}$

On integrating both sides,

$∫\frac{2ydy}{(y^2+1)}=-∫\frac{2xdx}{(x^2+1)}$

log(y²+ 1) = -log(x²+ 1) + log(c)

$(y^2+1)=[\frac{1}{(x^2+1)}]c$

$(y^2+1)=\frac{c}{(x^2+1)}$ (Where ‘c’ is integration constant)

Question 37 (ii). cosecx logy(dy/dx) + x²y²= 0

Solution:

We have,

cosecx logy(dy/dx) + x²y²= 0

log(y)dy/y²= -x²dx/cosecx

On integrating both sides,

∫[log(y)/y²]dy = -∫x²sinxdx

$log(y)∫\frac{dy}{y^2}-∫[\frac{d(logy)}{dy}∫\frac{dy}{y^2}]dy=-[x^2∫sinxdx-∫(\frac{d(x^2)}{dx}∫sinxdx)dx]+c$

-log(y)/y + ∫dy/y²= x²cosx – 2∫xcosxdx + c

-log(y)/y – 1/y = x²cosx – 2[x∫cosxdx – ∫{dx/dx∫cosxdx}dx] + c

-[{log(y) + 1}/y] = x²cosx – 2(xsinx – ∫sinxdx) + c

x²cosx + [{log(y) + 1}/y] – 2(xsinx + cosx) = c

Question 38 (i). xy(dy/dx) = 1 + x + y + xy

Solution:

We have,
xy(dy/dx) = 1 + x + y + xy
xy(dy/dx) = (1 + x) + y(1 + x)
xy(dy/dx) = (1 + x)(1 + y)
ydy/(1 + y) = [(1 + x)/x]dx
On integrating both sides,
∫ydy/(1 + y) = ∫[(1 + x)/x]dx
∫[1 – 1/(1 + y)]dy = ∫(dx/x) + ∫dx
y – log(1 + y) = log(x) + x + log(c)
y = log(x) + log(1 + y) + x + log(c)
y = log[cx(1 + y)] + x (Where ‘c’ is integration constant)

Question 38 (ii). y(1 – x²)(dy/dx) = x(1 + y²)

Solution:

We have,

y(1 – x²)(dy/dx) = x(1 + y²)

$\frac{ydy}{(1+y^2)}=\frac{xdx}{(1-x^2)}$

On integrating both sides,

$∫\frac{ydy}{(1+y^2)}=∫\frac{xdx}{(1-x^2)}$

Multiplying both sides by 2,

$∫\frac{2ydy}{(1+y^2)}=∫\frac{2xdx}{(1-x^2)}$

log(1 + y²) = -log(1 – x²) + log(c)

log[(1 + y²)(1 – x²)] = logc

(1 + y²)(1 – x²) = c (Where ‘c’ is integration constant)

Question 38 (iii). ye^x/ydx = (xe^x/y+ y²)dy

Solution:

We have,
ye^x/ydx = (xe^x/y+ y²)dy
ye^x/ydx – xe^x/ydy = y²dy
e^x/y(ydx – xdy)/y²= dy
e^x/yd(x/y) = dy
On integrating both sides,
∫e^x/yd(x/y) = ∫dy
e^x/y= y + c (Where ‘c’ is integration constant)

Question 38 (iv). (1 + y²)tan^-1xdx + 2y(1 + x²)dy = 0

Solution:

We have,

(1 + y²)tan^-1xdx + 2y(1 + x²)dy = 0 -(i)

$\frac{ydy}{(1+y^2)}=-[\frac{tan^{-1}x}{2(1+x^2)}]dx$

On integrating both sides,

$∫\frac{ydy}{(1+y^2)}=-∫[\frac{tan^{-1}x}{2(1+x^2)}]dx$ -(ii)

Let, I = $∫[\frac{tan^{-1}x}{2(1+x^2)}]dx$

$I=tan^{-1}x∫\frac{1}{2(x^2+1)}dx-∫[\frac{d}{dx}(tan^{-1}x)∫\frac{1}{2(x^2+1)}dx]dx$

$I=tan^{-1}x(\frac{1}{2}tan^{-1}x)-∫\frac{tan^{-1}x}{2(1+x^2)}dx$

$I=tan^{-1}x(\frac{1}{2}tan^{-1}x)-I$

2I = (1/2)(tan^-1x)²

I = (1/4)(tan^-1x)²

From equation (ii)

(1/2)log(1 + y²) = -(1/4)(tan^-1x)²+ c

log(1 + y²) + (1/2)(tan^-1x)²= c

Question 39. (dy/dx) = ytan2x, y(0) = 2

Solution:

We have,
(dy/dx) = ytan2x
(dy/y) = tan2xdx
On integrating both sides,
∫(dy/y) = ∫tan2xdx
log(y) = (1/2)log(sec2x) + log(c)
y = c(sec2x)^1/2
Put x = 0, y = 2 in above equation
c = 2
y = 2(sec2x)^1/2

Solve the differential equations(question 40-48):

Question 40. 2x(dy/dx) = 3y, y(1) = 2

Solution:

We have,
2x(dy/dx) = 3y
2dy/y = 3dx/x
On integrating both sides,
2∫dy/y = 3∫dx/x
2log(y) = 3log(x) + log(c)
y²= x³c
Put x = 1, y = 2 in above equation
c = 4
y²= 4x³

Question 41. xy(dy/dx) = y + 2, y(2) = 0

Solution:

We have,
xy(dy/dx) = y + 2
ydy/(y + 2) = dx/x
On integrating both sides,
∫ydy/(y + 2) = ∫(dx/x)
∫1 – \frac{2}{(y+2)} dy = ∫(dx/x)
y – 2log(y + 2) = log(x) + log(c)
Put x = 2, y = 0 in above equation
0 – 2log(2) = log(2) + log(c)
log(c) = -3log(2)
log(c) = log(1/8)
c = (1/8)
y – 2log(y + 2) = log(x/8)

Question 42. (dy/dx) = 2e^xy³, y(0) = 1/2

Solution:

We have,
(dy/dx) = 2e^xy³
dy/y³= 2e^xdx
On integrating both sides,
∫dy/y³= 2∫e^xdx
-(1/2y²) = 2e^x+ c
Put x = 0, y = (1/2) in above equation
-(4/2) = 2 + c
c = -4
-(1/2y²) = 2e^x– 4
y²(4e^x– 8) = -1
y²(8 – 4e^x) = 1

Question 43. (dr/dt) = -rt, r(0) = r₀

Solution:

We have,
(dr/dt) = -rt
dr/r = -tdt
On integrating both sides,
∫dr/r = -∫tdt
log(r) = -t²/2 + c
Put t = 0, r =r₀ in above equation
c = log(r₀)
log(r) = -t2/2 + log(r₀)
log(r/r₀) = -t²/2
(r/r₀) = $e^{-\frac{t^2}{2}}$
r = r₀ $e^{-\frac{t^2}{2}}$

Question 44. (dy/dx) = ysin2x, y(0) = 1

Solution:

We have,
(dy/dx) = ysin2x
dy/y = sin2xdx
On integrating both sides,
∫(dy/y) = ∫sin2xdx
log(y) = -(1/2)cos2x + c
Put x = 0, y = 1 in above equation
log|1| = -cos0/2 + c
c = (1/2)
log(y) = (1/2) – (cos2x/2)
log(y) = (1 – cos2x)/2
log(y) = 2sin²x/2
log(y) = sin²x
y = $e^{sin^2x}$

Question 45(i). (dy/dx) = ytanx, y(0) = 1

Solution:

We have,
(dy/dx) = ytanx
(dy/y) = tanxdx
On integrating both sides
∫(dy/y) = ∫tanxdx
log(y) = log(secx) + c
Put x = 0, y = 1 in above equation
0 = log(1) + c
c = 0
log(y) = log(secx)
y = secx

Question 45(ii). 2x(dy/dx) = 5y, y(1) = 1

Solution:

We have,
2x(dy/dx) = 5y
(2dy/y) = 5dx/x
On integrating both sides
2∫(dy/y) = 5∫(dx/x)
2log(y) = 5log(x) + c
Put x = 1, y = 1 in above equation
2log(1) = 5log(1) + c
c = 0
2log(y) = 5log(x)
y²= x⁵
y = |x|^(5/2)

Question 45(iii). (dy/dx) = 2e^2xy², y(0) = -1

Solution:

We have,
(dy/dx) = 2e^2xy²
(dy/y2) = 2e^2xdx
On integrating both sides
∫(dy/y²) = 2∫e^2xdx
-(1/y) = 2e^2x/2 + c
-(1/y) = e^2x+ c
Put x = 0, y = -1 in above equation
1 = e⁰+ c
c = 0
-(1/y) = e^2x
y = -e^-2x

Question 45(iv). cosy(dy/dx) = e^x, y(0) = π/2

Solution:

We have,
cosy(dy/dx) = e^x
cosydy = e^xdx
On integrating both sides
∫cosydy = ∫e^xdx
siny = e^x+ c
Put x = 1, y = π/2 in above equation
sin(π/2) = e⁰+ c
1 = 1 + c
c = 0
siny = e^x
y = sin^-1(e^x)

Question 45(v). (dy/dx) = 2xy, y(0) = 1

Solution:

We have,
(dy/dx) = 2xy
dy/y = 2xdx
On integrating both sides
∫(dy/y) = 2∫xdx
log(y) = x²+ c
Put x = 0, y = 1 in above equation
log(1) = 0 + c
c = 0
log(y) = x²

Question 45(vi). (dy/dx) = 1 + x²+ y²+ x²y², y(0) = 1

Solution:

We have,

(dy/dx) = 1 + x²+ y²+ x²y²

(dy/dx) = (1 + x²) + y²(1 + x²)

(dy/dx) = (1 + x²)(1 + y²)

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

On integrating both sides

$∫\frac{dy}{(1+y^2)}=∫(1+x^2)dx$

tan^-1y = x + (x³/3) + c

Put x = 0, y = 1 in above equation

tan^-1(1) = 0 + 0 + c

c = π/4

tan^-1y = x + (x³/3) + π/4

Question 45(vii). xy(dy/dx) = (x + 2)(y + 2), y(1) = -1

Solution:

We have,
xy(dy/dx) = (x + 2)(y + 2)
ydy/(y + 2) = (x + 2)dx/x
On integrating both sides
∫ydy/(y + 2) = ∫(x + 2)dx/x
∫[1 – 2/(y + 2)]dy = ∫dx + 2∫(dx/x)
y – 2log(y + 2) = x + 2log(x) + c
Put x = 1, y = -1 in above equation
-1 – 2log(-1 + 2) = 1 + 2log(1) + c
c = -2
y – 2log(y + 2) = x + 2log(x) – 2

Question 45(viii). (dy/dx) = 1 + x + y²+ xy², y(0) = 0

Solution:

We have,

(dy/dx) = 1 + x + y²+ xy²

(dy/dx) = (1 + x) + y²(1 + x)

(dy/dx) = (1 + x)(1 + y²)

On integrating both sides

$∫\frac{dy}{(1+y^2)}=∫(1+x)dx$

tan^-1(y) = x + (x²/2) + c

Put x = 0, y = 0 in above equation

tan^-1(0) = 0 + 0 + c

c = 0

tan^-1(y) = x + (x²/2)

y = tan(x + x²/2)

Question 45(ix). 2(y + 3) – xy(dy/dx) = 0, y(1) = -2

Solution:

We have,
2(y + 3) – xy(dy/dx) = 0
xy(dy/dx) = 2(y + 3)
ydy/(y + 3) = 2(dx/x)
On integrating both sides
∫[ydy/(y + 3)] = 2∫(dx/x)
∫[1 – 3/(y + 3)]dy = 2∫(dx/x)
y – 3log(y + 3) = 2log(x) + c
Put x = 1, y = -2 in above equation
-2 – 3log(-2 + 3) = 2log(1) + c
c = -2
y – 3log(y + 3) = 2log(x) – 2
y + 2 = log(x)²log(y + 3)³
e^(y+2)= x²(y + 3)³

Question 46. x(dy/dx) + coty = 0, y = π/4 at x = √2

Solution:

We have,
x(dy/dx) + coty = 0
x(dy/dx) = -coty
dy/coty = -dx/x
On integrating both sides
∫dy/coty = -∫dx/x
∫tanydy = -∫(dx/x)
log(secy) = -log(x) + c
log(xsecy) = c
Put x = √2, y = π/4 in above equation
log|√2.√2| = c
c = log(2)
log(xsecy) = log(2)
x/cosy = 2
x = 2cosy

Question 47. (1 + x²)(dy/dx) + (1 + y²) = 0, y = 1 at x = 0

Solution:

We have,

(1 + x²)(dy/dx) + (1 + y²) = 0

(1 + x²)(dy/dx) = -(1 + y²)

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

On integrating both sides

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

tan^-1y = -tan^-1x + c

Put x = 0, y = 1 in above equation

tan^-1(1) = tan^-1(0) + c

c = π/4

tan^-1y = π/4 – tan^-1x

y = tan(π/4 – tan^-1x)

$y=\frac{tan\frac{π}{4}-tan(tan^{-1}x)}{1+tan\frac{π}{4}tan(tan^{-1}x)}$

y = (1 – x)/(1 + x)

y + yx = 1 – x

x + y = 1 – xy

Question 48. (dy/dx) = 2x(logx + 1)/(siny + ycosy), y = 0 at x = 1

Solution:

We have,
(dy/dx) = 2x(logx + 1)/(siny + ycosy)
(siny + ycosy)dy = 2x(logx + 1)dx
On integrating both sides
∫sinydy + ∫ycosydy = 2∫xlogxdx + 2∫xdx
-cosy + y∫cosydy – ∫[(dy/dy)∫cosydy]dy = 2logx∫xdx – 2∫[( $\frac{d(logx)}{dx}$ ∫xdx]dx + x²+ c
-cosy + ysiny – ∫sinydy = x²logx – ∫xdx + x²+ c
-cosy + ysiny + cosy = x²logx – x²/2 + x²+ c
Put x = 1, y = 0 in above equation
-1 + 0 + 1 = 0 – (1/2) + 1 + c
c = -(1/2)
ysiny = x²logx + x²/2 – (1/2)
2ysiny = 2x²logx + x²– 1

Question 49. Find the particular solution of the differential equation e^(dy/dx)= x + 1, given that y(0) = 3 when x = 0.

Solution:

We have,
e^(dy/dx)= x + 1
Taking log both sides,
(dy/dx) = log(x + 1)
dy = log(x + 1)dx
On integrating both sides
∫dy = ∫log(x + 1)dx
y = log(x + 1)∫dx – ∫[ $\frac{d(log(x+1))}{dx}$ ∫dx]dx
y = xlog(x + 1) – ∫xdx/(x + 1)
y = xlog(x + 1) – ∫[1 – 1/(x + 1)]dx
y = xlog(x + 1) – x + log(x + 1) + c
y = (x + 1)log(x + 1) – x + c
Put x = 0, y = 3 in above equation
3 = 0 – 0 + c
c = 3
y = (x + 1)log(x + 1) – x + 3

Question 50. Find the solution of the differential equation cosydy + cosxsinydx = 0, given that y = π/2 when x = π/2.

Solution:

We have,
cosydy + cosxsinydx = 0
cosydy = -cosxsinydx
(cosy/siny)dy = -cosxdx
On integrating both sides
∫cotydy = -∫cosxdx
log(siny) = -sinx + c
Put x = π/2, y = π/2 in above equation
log|sinπ/2| = -sin(π/2) + c
0 = -1 + c
c = 1
log(siny) = 1 – sin(x)
log(siny) + sin(x) = 1

Question 51. Find the particular solution of the differential equation (dy/dx) = -4xy², given that y =1 when x = 0.

Solution:

We have,
(dy/dx) = -4xy²
(dy/y²) + 4xdx = 0
On integrating both sides
∫(dy/y²) + 4∫xdx = 0
-(1/y) + 2x²= c
Put x = 0, y = 1 in above equation
-1 + 0 = c
c = -1
-(1/y) + 2x²= -1
(1/y) = 2x²+ 1
y = 1/(2x²+ 1)

Question 52. Find the equation of a curve passing through the point(0, 0) and whose differential equation is (dy/dx) = e^xsinx.

Solution:

We have,
(dy/dx) = e^xsinx
dy = e^xsinxdx
On integrating both sides
∫dy = ∫e^xsinxdx
Let, I = ∫e^xsinxdx
I = e^x∫sinx – ∫[ $\frac{de^x}{dx}$ ∫sinxdx]dx
I = -e^xcosx + ∫e^xcosxdx
I = -e^xcosx + e^x∫cosxdx – ∫[ $\frac{de^x}{dx}$ ∫cosxdx]dx
I = -e^xcosx + e^xsinx – ∫e^xsinxdx
I = -e^xcosx + e^xsinx – I
2I = -e^xcosx + e^xsinx
I = e^x(sinx – cosx)/2
y = e^x(sinx – cosx)/2

Question 53. For the differential equation xy(dy/dx) = (x + 2)(y + 2), find the solution curve passing through the point (1, -1).

Solution:

We have,
xy(dy/dx) = (x + 2)(y + 2)
ydy/(y + 2) = (x + 2)dx/x
On integrating both sides
∫ydy/(y + 2) = ∫(x + 2)dx/x
∫[1 – 2/(y + 2)]dy = ∫dx + 2∫(dx/x)
y – 2log(y + 2) = x + 2log(x) + c
y – x – c = log(x)²+ log(y + 2)²
y – x – c = log|x²(y + 2)²|
Curve is passing through (1, -1)
-1 – 1 – c = log(1)
c = 2
y – x – 2 = log|x²(y + 2)²|

Question 54. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Solution:

We have,

Let, v be the volume of the sphere, t be the time, r be the radius of sphere & k is a constant

Volume of sphere is given by v = (4/3)πr³

According to the question (dv/dt) = k

$\frac{d[(4/3)πr^3]}{dt} = k$

(4/3)π.3r²(dr/dt) = k

4πr²dr = kdt

On integrating both sides

∫4πr²dr = ∫kdt

4π(r³/3) = kt + c

4πr³= 3(kt + c) -(i)

At t = 0, r = 38

4π(3)³= 3(0 + c)

c = 36π

At t = 3, r = 6 in equation (i)

4π(6)³= 3(kt + 36π)

864π = 9k + 108π

k = 84π

4πr³= 3(84πt + 36π)

r³= 63t + 27

r = (63t + 27)^1/3

Radius of the balloon after t second is (63t + 27)^1/3

Question 55. In a bank principal increases at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log 2 = 0.6931).

Solution:

We have,
Let ‘p’ and ‘t’ be the principal and time respectively.
Principal increases at the rate of r % per year.
dp/dt = (r/100)p
(dp/p) = (r/100)dt
On integrating both sides
∫(dp/p) = (r/100)∫dt
log(p) = (rt/100) + c -(i)
At t = 0, p = 100
log(100) = 0 + c
c = log(100) -(ii)
If t = 10, p = 2 × 100 in equation (i)
log(200) = (10r/100) + log(100)
log(200/100) = (10r/100)
log(2) = (r/10)
0.6931 = (r/10)
r = 6.931

Question 56. In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e = 1.648).

Solution:

We have,
Let ‘p’ and ‘t’ be the principal and time respectively.
Principal increases at the rate of 5% per year,
(dp/dt) = (5/100)p -(i)
(dp/p) = (1/20)dt
On integrating both sides
∫(dp/p) = (1/20)∫dt
log(p) = (t/20) + c -(ii)
At t = 0, p = 1000
log(1000) = c
log(p) = (t/20) + log(1000)
Putting t = 10 in equation in (i)
log(p/1000) = (10/20)
p = 1000e^0.5
p = 1000 × 1.648
p = 1648

Question 57. In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Solution:

We have,

Let numbers of bacteria at time ‘t’ be ‘x’

The rate of growth of bacteria is proportional to the number present

(dx/dt)∝ x -(i)

(dx/dt) = kx (where ‘k’ is proportional constant)

(dx/x) = kdt

On integrating both sides

∫(dx/x) = k∫dt

log(x) = kt + c -(ii)

At t = 0, x = x₀(x₀ is numbers of bacteria at t = 0)

log(x₀) = 0 + c

c = log(x₀)

On putting the value of c in equation (ii)

log(x) = kt + log(x₀)

log(x/x₀) = kt -(iii)

The number is increased by 10% in 2 hours.

x = x₀(1 + 10/100)

(x/x₀) = (11/10)

On putting the value of (x/x₀) & t = 2 in equation (iii)

2 × k = log(11/10)

k = (1/2)log(11/10)

Therefore, equation (iii) becomes

log(x/x₀) = (1/2)log(11/10) × t

$t=\frac{2log\frac{x}{x0}}{log\frac{11}{10}}$

At time t₁ numbers of bacteria becomes 200000 from 100000(i.e, x = 2x₀)

t₁ $=\frac{2log\frac{2x0}{x0}}{log\frac{11}{10}}$

t₁ $=\frac{2log2}{log\frac{11}{10}}$

Question 58. If y(x) is a solution of the differential equation $[\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx$ , and y(0) = 1, then find the value of y(π/2).

Solution:

We have,
$[\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx$ (i)
dy/(1 + y) = -[(cosx)/(2 + sinx)]dx
On integrating both sides
∫dy/(1 + y) = -∫[(cosx)/(2 + sinx)]dx
log(1 + y) = -log(2 + sinx) + log(c)
log(1 + y) + log(2 + sinx) = log(c)
(1 + y)(2 + sinx) = c
Put at x = 0, y = 1
c = (1 + 1)(2 + 0)
c = 4
(1 + y)(2 + sinx) = 4
(1 + y) = 4/(2 + s inx)
y = 4/(2 + sinx) – 1
We need to find the value of y(π/2)
y = 4/(2 + sinπ/2) – 1
y = (4/3) – 1
y = (1/3)

Solve the following:

Question 1: dy/dx = (x + y + 1)²

Solution:

We have,
dy/dx = (x + y + 1)²
Putting x + y + 1 = v
Therefore, dv/dx – 1 = v²
⇒ dv/dx = v²+ 1
⇒ 1/(v²+ 1) dv = dx
Integrating both sides, we get
∫ 1/(v²+ 1) dv = ∫ dx

Question 2: dy/dx cos (x – y) = 1

Solution:

We have,
dy/dx cos (x – y) = 1
⇒ dy/dx = 1/cos(x – y)
Putting x – y = v
⇒ 1 – dy/dx = dv/dx
⇒ dy/dx = 1 – dv/dx
Therefore, 1 – dv/dx = 1/cos v
⇒ dv / dx = 1 – 1/cos v
⇒ dv/dx = (cos v – 1)/cos v
⇒ cos v/ (cos v – 1) dv = dx
Integrating both sides, we get
∫cos v/(cos v – 1) dv = ∫dx
⇒ -∫(cos v (1 + cosv)) / (1 – cos² v) dv =∫ dx
⇒ -∫(cos v (1 + cos v)) / (sin² v) dv = ∫ dx
⇒ -∫(cot v cosec v + cot² v) dv = ∫ dx
⇒ -∫ (cot v cosec v + cosec² v – 1) dv = ∫ dx
⇒ -(-cosec v – cot v – v)= x + C
⇒ cosec ( x – y ) + cot ( x – y ) + x – y = x + C
⇒ cosec ( x – y ) + cot ( x – y ) – y = C
⇒ ((1+cos ( x – y )) / sin ( x – y )) – y = C
⇒ cot (( x – y )/ 2) = y + C

Question 3: dy/dx = ((x – y) + 3)/ (2(x – y) + 5)

Solution:

We have,
dy/dx = ((x – y) + 3)/ (2(x – y) + 5)
Putting x – y = v
⇒ 1 – dy/dx = dv/dx
⇒ dy/dx = 1 – dv/dx
Therefore, 1 – dv/dx = (v + 3)/ (2v + 5)
⇒ dv/dx = 1 – (v + 3)/ (2v + 5)
⇒ dv/dx = (2v + 5 – v – 3)/ 2v + 5
⇒ dv/dx = (v + 2) / (2v + 5)
⇒ (2v + 5)/(v + 2)dv = dx
Integrating both sides, we get
∫(2v + 5)/(v + 2) dv = ∫dx
⇒ ∫(2v + 4 + 1)/(v + 2) dv = ∫dx
⇒ ∫((2v + 4)/(v + 2) + 1/(v + 2))dv = ∫dx
⇒ 2∫dv + ∫1/(v + 2)dv = ∫dx
⇒ 2v + log |v + 2| = x + C
⇒ 2(x – y) + log | x – y + 2 | = x + C

Question 4: dy/dx = (x + y)²

Solution:

We have,
dy/dx = (x + y)²
Let x + y = v
⇒ 1 + dy/dx = dv/dx
⇒ dy/dx = dv/dx – 1
Therefore, dv/dx – 1 = v²
⇒ dv/dx = v² + 1
⇒ 1/(v² + 1) dv = dx
Integrating both sides, we get
∫1/(v² + 1) dv = ∫dx
⇒ tan^-1 v = x + C
⇒ v = tan (x + C)
⇒ x + y = tan (x + C)

Question 5: (x + y)² dy/dx = 1

Solution:

We have,
(x + y)² dy/dx = 1
⇒ dy/dx = 1/( x + y)²
Let x + y = v
⇒ 1 + dy/dx = dv/dx
⇒ dy/dx = dv/dx – 1
Therefore, dv/dx – 1 = 1/v²
⇒ dv/dx = 1/v² + 1
⇒ v²/(v² + 1) dv = dx
Integrating both sides, we get
∫v²/(v² + 1) dv = ∫dx
⇒ ∫v² + 1 – 1/(v² + 1) dv = ∫dx
⇒ ∫(1- 1/(v² + 1) dv = ∫dx
⇒ v – tan^-1 v = x + C
⇒ x + y – tan^-1(x + y) = x + C
⇒ y – tan^-1 (x + y) = C

Question 6: cos² ( x – 2y) = 1 – 2dy/dx

Solution:

We have,
cos² ( x – 2y ) = 1 – 2dy/dx
⇒ 2dy/dx = 1 – cos² (x – 2y)
Let x – 2y = v
⇒ 1 – 2 dy/dx = dv/dx
⇒ 2 dy/dx = 1 – dv/dx
Therefore, 1 – dv/dx = 1 – cos² v
⇒ dv/dx = cos² v
⇒ sec² v dv = dx
Integrating both sides, we get
∫ sec² v dv = ∫dx
⇒ tan v = x – C
⇒ tan (x – 2y) = x – C
⇒ x = tan (x – 2y) + C

Question 7: dy/dx = sec(x + y)

Solution:

We have,
dy/dx = sec(x + y)
⇒ dy/dx = 1/cos ( x + y)
Let x + y = v
⇒ 1 + dy/dx = dv/dx
⇒ dy/dx = dv/dx -1
Therefore, dv/dx – 1 = 1/cos v
⇒ dv/dx = (cos v + 1)/ cos v
⇒ cos v/(cos v + 1) dv = dx
Integrating both sides, we get
∫ cos v/(cos v + 1) dv = ∫ dx
⇒ ∫ cos v (1 – cos v)/(1 – cos² v ) dv = ∫ dx
⇒ ∫ cos v (1 – cos v)/sin² v dv = ∫ dx
⇒ ∫ (cos v – cos^2 v)/sin² v dv = ∫ dx
⇒ ∫(cot v cosec v – cot² v) dv = ∫ dx
⇒ ∫(cot v cosec v – cosec² v + 1) dv = ∫ dx
⇒ – cosec v + cot v + v = x + C
⇒ – cosec (x + y) + cot (x + y) + x + y = x + C
⇒ – cosec (x + y) + cot (x + y) + y = C
⇒ ((-1 + cos (x + y)) / sin (x + y)) + y = C
⇒ – tan ((x + y)/2) + y = C
⇒ y = tan((x + y)/2) + C

Question 8: dy/dx = tan (x + y)

Solution:

We have,
dy/dx = tan (x + y)
dy/dx = sin (x + y)/cos (x + y)
Let x + y =v
Therefore, 1 + dy/dx = dv/dx
⇒ dy/dx = dv/dx – 1
Since, dv/dx -1 = sin v/cos v
⇒ dy/dx = sin v/cos v + 1
⇒ dy/dx = (sin v + cos v)/ cos v
⇒ cos v/(sin v + cos v) dv = dx
Integrating both sides, we get
⇒ ∫ cos v/(sin v + cos v) dv =∫ dx
⇒ 1/2 ∫ {(sin v + cos v) + (cos v – sin v)}/(sin v + cos v) dv = ∫dx
⇒ 1/2 ∫ dv + 1/2 ∫ (cos v – sin v) / (sin v + cos v) dv = ∫dx
⇒ 1/2 v + 1/2 ∫ (cos v – sin v)/(sin v + cos v) dv = x
Putting sin v + cos v = t
⇒ (cos v – sin v) dv = dt
Therefore, 1/2 v + 1/2 ∫ dt/t = x
⇒ 1/2 v + 1/2 log |t| = x + C
⇒ (x + y) + 1/2 log |sin (x + y) + cos (x + y)| = x + C
⇒ 1/2 (y – x) + 1/2 log |sin (x + y) + cos (x + y)| = C
⇒ (y – x) + log |sin (x + y) + cos (x + y)| = 2C
⇒ y – x + log |sin (x + y) + cos (x + y)| = K (where K = 2C)

Question 9: (x + y) (dx – dy) = dx + dy

Solution:

We have,
(x + y) (dx – dy) = dx + dy
⇒ x dx + y dx -x dy – y dy = dx + dy
⇒ (x + y -1)dx = ( x + y +1) dy
⇒ dy/dx = (x + y -1)/(x + y + 1)
Let x + y = v
Therefore, 1 + dy/dx =dv/dx
⇒ dy/dx = dv/dx – 1
Therefore, dv/dx -1 = (v – 1)/(v +1)
⇒ dv/dx = (v – 1)/(v +1) + 1
⇒ dv/dx = (v – 1 + v +1)/(v +1)
⇒ dv/dx = 2v / (v +1)
⇒ (v +1) / 2v dv = dx
Integrating both sides, we get
∫(v + 1)/2v dv = ∫ dx
⇒ 1/2 ∫dv + 1/2 ∫1/v dv = ∫dx
⇒ 1/2 v + 1/2 log |v| = x + C
⇒ 1/2 (x + y) + 1/2 log |x + y| = x + C
⇒ 1/2 (y – x) + 1/2 log |x + y| = C

Question 10: (x + y + 1)dy/dx = 1

Solution:

We have,
(x + y + 1)dy/dx =1
⇒ dy/dx = 1/(x + y + 1)
Let x + y + 1 = v
Therefore, 1 + dy/dx = dv/dx
⇒ dy/dx = dv/dx – 1
Therefore, dv/dx – 1= 1/v
⇒ dv/dx = 1/v + 1
⇒ v/(v + 1) dv = dx
Integrating both sides, we get
∫ v/(v + 1) dv = ∫ dx
⇒ ∫ (v + 1 – 1)/(v + 1) dv = ∫ dx
⇒ ∫ (1 – 1/(v + 1))dv = ∫ dx
⇒ v – log |v + 1| = x + K
⇒ x + y + 1 – log |x + y+ 1 + 1| = x + K
⇒ y – log |x + y + 2| = K – 1
⇒ y – log |x + y + 2| = C₁ ( C₁ = K – 1)
⇒ y – C₁ = log |x + y + 2|
⇒ e^{y – C}_¹ = x + y + 2
⇒ e^y/ e^C_¹ = x + y + 2
⇒ e^{– C}_¹ e^y = x + y + 2
⇒ C e^y = x + y + 2 (C = e^{– C}_¹)
⇒ x = C e^y – y – 2

Question 11: dy/dx + 1 = e^{x + y}

Solution:

We have,
dy/dx + 1 = e^{x + y} . . . (1)
Let x + y = t
⇒ 1 + dy/dx = dt/dx
Substituting the value of x + y = t and 1 + dy/dx = dt/dx from (1), we get
dt/dx = e^t
⇒ e^{– t} dt = dx
⇒ – e^{– t} = x + C
⇒ – e^{– (x + y)} = x + C [Since t = x + y]

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RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations

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)

Solve the following differential equations:

Question 21. (1 – x2)dy + xydx = xy2dx

Question 22. tanydx + sec2ytanxdy = 0

Question 23. (1 + x)(1 + y2)dx + (1 + y)(1 + x2)dy = 0

Question 24. tany(dy/dx) = sin(x + y) + sin(x – y)

Question 25. cosxcosy(dy/dx) = -sinxsiny

Question 26. (dy/dx) + cosxsiny/cosy = 0

Question 27. x√(1 – y2)(dx) + y√(1 – x2)dy = 0

Question 28. y(1 + ex)dy =(y + 1)exdx

Question 29. (y + xy)dx + (x – xy2)dy = 0

Question 30. (dy/dx) = 1 – x + y – xy

Question 31. (y2 + 1)dx – (x2 + 1)dy = 0

Question 32. dy + (x + 1)(y + 1)dx = 0

Question 33. (dy/dx) = (1 + x2)(1 + y2)

Question 34. (x – 1)(dy/dx) = 2x3y

Question 35. (dy/dx) = ex+y + e-x+y

Question 36. (dy/dx) = (cos2x – sin2x)cos2y

Question 37(i). (xy2 + 2x)(dx) + (x2y + 2y)dy = 0

Question 37 (ii). cosecx logy(dy/dx) + x2y2 = 0

Question 38 (i). xy(dy/dx) = 1 + x + y + xy

Question 38 (ii). y(1 – x2)(dy/dx) = x(1 + y2)

Question 38 (iii). yex/ydx = (xex/y + y2)dy

Question 38 (iv). (1 + y2)tan-1xdx + 2y(1 + x2)dy = 0

Question 39. (dy/dx) = ytan2x, y(0) = 2

Solve the differential equations(question 40-48):

Question 40. 2x(dy/dx) = 3y, y(1) = 2

Question 41. xy(dy/dx) = y + 2, y(2) = 0

Question 42. (dy/dx) = 2exy3, y(0) = 1/2

Question 43. (dr/dt) = -rt, r(0) = r0

Question 44. (dy/dx) = ysin2x, y(0) = 1

Question 45(i). (dy/dx) = ytanx, y(0) = 1

Question 45(ii). 2x(dy/dx) = 5y, y(1) = 1

Question 45(iii). (dy/dx) = 2e2xy2, y(0) = -1

Question 45(iv). cosy(dy/dx) = ex, y(0) = π/2

Question 45(v). (dy/dx) = 2xy, y(0) = 1

Question 45(vi). (dy/dx) = 1 + x2 + y2 + x2y2, y(0) = 1

Question 45(vii). xy(dy/dx) = (x + 2)(y + 2), y(1) = -1

Question 45(viii). (dy/dx) = 1 + x + y2 + xy2, y(0) = 0

Question 45(ix). 2(y + 3) – xy(dy/dx) = 0, y(1) = -2

Question 46. x(dy/dx) + coty = 0, y = π/4 at x = √2

Question 47. (1 + x2)(dy/dx) + (1 + y2) = 0, y = 1 at x = 0

Question 48. (dy/dx) = 2x(logx + 1)/(siny + ycosy), y = 0 at x = 1

Question 49. Find the particular solution of the differential equation e(dy/dx) = x + 1, given that y(0) = 3 when x = 0.

Question 50. Find the solution of the differential equation cosydy + cosxsinydx = 0, given that y = π/2 when x = π/2.

Question 51. Find the particular solution of the differential equation (dy/dx) = -4xy2, given that y =1 when x = 0.

Question 52. Find the equation of a curve passing through the point(0, 0) and whose differential equation is (dy/dx) = exsinx.

Question 53. For the differential equation xy(dy/dx) = (x + 2)(y + 2), find the solution curve passing through the point (1, -1).

Question 54. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Question 55. In a bank principal increases at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log 2 = 0.6931).

Question 56. In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e = 1.648).

Question 57. In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Question 58. If y(x) is a solution of the differential equation , and y(0) = 1, then find the value of y(π/2).

Solve the following:

Question 1: dy/dx = (x + y + 1)2

Question 2: dy/dx cos (x – y) = 1

Question 3: dy/dx = ((x – y) + 3)/ (2(x – y) + 5)

Question 4: dy/dx = (x + y)2

Question 5: (x + y)2 dy/dx = 1

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

Question 21. (1 – x²)dy + xydx = xy²dx

Question 22. tanydx + sec²ytanxdy = 0

Question 23. (1 + x)(1 + y²)dx + (1 + y)(1 + x²)dy = 0

Question 27. x√(1 – y²)(dx) + y√(1 – x²)dy = 0

Question 28. y(1 + e^x)dy =(y + 1)e^xdx

Question 29. (y + xy)dx + (x – xy²)dy = 0

Question 31. (y²+ 1)dx – (x²+ 1)dy = 0

Question 33. (dy/dx) = (1 + x²)(1 + y²)

Question 34. (x – 1)(dy/dx) = 2x³y

Question 35. (dy/dx) = e^x+y+ e^-x+y

Question 36. (dy/dx) = (cos²x – sin²x)cos²y

Question 37(i). (xy²+ 2x)(dx) + (x²y + 2y)dy = 0

Question 37 (ii). cosecx logy(dy/dx) + x²y²= 0

Question 38 (ii). y(1 – x²)(dy/dx) = x(1 + y²)

Question 38 (iii). ye^x/ydx = (xe^x/y+ y²)dy

Question 38 (iv). (1 + y²)tan^-1xdx + 2y(1 + x²)dy = 0

Question 42. (dy/dx) = 2e^xy³, y(0) = 1/2

Question 43. (dr/dt) = -rt, r(0) = r₀

Question 45(iii). (dy/dx) = 2e^2xy², y(0) = -1

Question 45(iv). cosy(dy/dx) = e^x, y(0) = π/2

Question 45(vi). (dy/dx) = 1 + x²+ y²+ x²y², y(0) = 1

Question 45(viii). (dy/dx) = 1 + x + y²+ xy², y(0) = 0

Question 47. (1 + x²)(dy/dx) + (1 + y²) = 0, y = 1 at x = 0

Question 49. Find the particular solution of the differential equation e^(dy/dx)= x + 1, given that y(0) = 3 when x = 0.

Question 51. Find the particular solution of the differential equation (dy/dx) = -4xy², given that y =1 when x = 0.

Question 52. Find the equation of a curve passing through the point(0, 0) and whose differential equation is (dy/dx) = e^xsinx.

Question 58. If y(x) is a solution of the differential equation $[\frac{(2 + sinx)}{(1+y)}](\frac{dy}{dx})=-cosx$ , and y(0) = 1, then find the value of y(π/2).

Question 1: dy/dx = (x + y + 1)²

Question 4: dy/dx = (x + y)²

Question 5: (x + y)² dy/dx = 1

Question 6: cos² ( x – 2y) = 1 – 2dy/dx

Question 11: dy/dx + 1 = e^{x + y}