Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.7 | Set 2

Last Updated : 05 Mar, 2021

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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

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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.7 | Set 2

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

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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