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

### Solve the following differential equations:

### Question 21. (1 – x^{2})dy + xydx = xy^{2}dx

**Solution:**

We have,

(1 – x

^{2})dy + xydx = xy^{2}dx(1 – x

^{2})dy = xy^{2}dx – xydx(1 – x

^{2})dy = xy(y – 1)dxOn integrating both sides,

log(y – 1) – logy = -(1/2)log(1 – x

^{2}) + logclog(y – 1) – logy + (1/2)log(1 – x

^{2}) = logc (Where ‘c’ is integration constant)

### Question 22. tanydx + sec^{2}ytanxdy = 0

**Solution:**

We have,

tanydx + sec

^{2}ytanxdy = 0tanydx = -sec

^{2}ytanxdy(sec

^{2}y/tany)dy = -dx/tanxOn integrating both sides,

∫(sec2y/tany)dy = -∫cotxdx

Let, tany = z

On differentiating both sides

sec

^{2}xdx = 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 + y^{2})dx + (1 + y)(1 + x^{2})dy = 0

**Solution:**

We have,

(1 + x)(1 + y

^{2})dx + (1 + y)(1 +x^{2})dy = 0On integrating both sides,

tan

^{-1}(y) + (1/2)log(1 + y^{2}) = -tan^{-1}(x) – (1/2)log(1 + x^{2}) + ctan

^{-1}(y) + tan^{-1}(x) + (1/2)log[(1 + y^{2})(1 + x^{2})] = 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 – y^{2})(dx) + y√(1 – x^{2})dy = 0

**Solution:**

We have,

x√(1 – y

^{2})(dx) + y√(1 – x^{2})dy = 0x√(1 – y

^{2})(dx) = -y√(1 – x^{2})dyOn integrating both sides,

√(1 – y

^{2}) = -√(1 – x^{2}) + c√(1 – y

^{2}) + √(1 – x^{2}) = c (Where ‘c’ is integration constant)

### Question 28. y(1 + e^{x})dy =(y + 1)e^{x}dx

**Solution:**

We have,

y(1 + e

^{x})dy =(y + 1)e^{x}dxOn integrating both sides,

∫[1 – 1/(y + 1)]dy = ∫e

^{x}dx/(1 + e^{x})y – log(y + 1) = log(1 + e

^{x}) + c (Where ‘c’ is integration constant)

### Question 29. (y + xy)dx + (x – xy^{2})dy = 0

**Solution:**

We have,

(y + xy)dx + (x – xy

^{2})dy = 0y(1 + x)dx = -x(1 – y

^{2})dy[(1 – y

^{2})/y]dy = -[(1 + x)/x]dxOn integrating both sides,

∫[(1 – y

^{2})/y]dy = -∫[(1 + x)/x]dx∫(dy/y) – ∫ydy = -∫dx/x – ∫dx

log(y) – (y

^{2}/2) = -log(x) – x + clog(x) + x + log(y) – (y

^{2}/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 – (x

^{2}/2) + c (Where ‘c’ is integration constant)

### Question 31. (y^{2 }+ 1)dx – (x^{2 }+ 1)dy = 0

**Solution:**

We have,

(y

^{2 }+ 1)dx – (x^{2 }+ 1)dy = 0(y

^{2 }+ 1)dx = (x^{2 }+ 1)dyOn integrating both sides,

tan

^{-1}y = tan^{-1}x + 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) = -(x

^{2}/2) – x + clog(y + 1) + (x

^{2}/2) + x = c (Where ‘c’ is integration constant)

### Question 33. (dy/dx) = (1 + x^{2})(1 + y^{2})

**Solution:**

We have,

(dy/dx) = (1 + x

^{2})(1 + y^{2})On integrating both sides,

tan

^{-1}y = x + (x^{3}/3) + ctan

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

### Question 34. (x – 1)(dy/dx) = 2x^{3}y

**Solution:**

We have,

(x – 1)(dy/dx) = 2x

^{3}ydy/y = 2x

^{3}dx/(x – 1)On integrating both sides,

∫dy/y = 2∫x

^{3}dx/(x – 1)log(y) = (2/3)(x

^{3}) + 2(x^{2}/2) + 2x + 2log(x – 1) + log(c)y = c|x – 1|

^{2}e^{[(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})dxOn integrating both sides,

∫e

^{-y}dy = ∫e^{x}dx + ∫e^{-x}dx-e

^{-y }= e^{x }– e^{-x }+ ce

^{-x}-e^{-y }= e^{x }+ c (Where ‘c’ is integration constant)

### Question 36. (dy/dx) = (cos^{2}x – sin^{2}x)cos^{2}y

**Solution:**

We have,

(dy/dx) = (cos

^{2}x – sin^{2}x)cos^{2}ydy/cos

^{2}y = (cos^{2}x – sin^{2}x)dxsex

^{2}ydy = cos2xdxOn integrating both sides,

∫sex

^{2}ydy = ∫cos2xdxtany = (sin2x/2) + c (Where ‘c’ is integration constant)

### Question 37(i). (xy^{2 }+ 2x)(dx) + (x^{2}y + 2y)dy = 0

**Solution:**

We have,

(xy

^{2 }+ 2x)(dx) + (x^{2}y + 2y)dy = 0x(y

^{2 }+ 2)(dx) = -y(x^{2 }+ 2)dyMultiplying both sides by 2,

On integrating both sides,

log(y

^{2 }+ 1) = -log(x^{2 }+ 1) + log(c)(Where ‘c’ is integration constant)

### Question 37 (ii). cosecx logy(dy/dx) + x^{2}y^{2 }= 0

**Solution:**

We have,

cosecx logy(dy/dx) + x

^{2}y^{2 }= 0log(y)dy/y

^{2 }= -x^{2}dx/cosecxOn integrating both sides,

∫[log(y)/y

^{2}]dy = -∫x^{2}sinxdx-log(y)/y + ∫dy/y

^{2 }= x^{2}cosx – 2∫xcosxdx + c-log(y)/y – 1/y = x

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

^{2}cosx – 2(xsinx – ∫sinxdx) + cx

^{2}cosx + [{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 – x^{2})(dy/dx) = x(1 + y^{2})

**Solution:**

We have,

y(1 – x

^{2})(dy/dx) = x(1 + y^{2})On integrating both sides,

Multiplying both sides by 2,

log(1 + y

^{2}) = -log(1 – x^{2}) + log(c)log[(1 + y

^{2})(1 – x^{2})] = logc(1 + y

^{2})(1 – x^{2}) = c (Where ‘c’ is integration constant)

### Question 38 (iii). ye^{x/y}dx = (xe^{x/y }+ y^{2})dy

**Solution:**

We have,

ye

^{x/y}dx = (xe^{x/y }+ y^{2})dyye

^{x/y}dx – xe^{x/y}dy = y^{2}dye

^{x/y}(ydx – xdy)/y^{2 }= dye

^{x/y}d(x/y) = dyOn integrating both sides,

∫e

^{x/y}d(x/y) = ∫dye

^{x/y }= y + c (Where ‘c’ is integration constant)

### Question 38 (iv). (1 + y^{2})tan^{-1}xdx + 2y(1 + x^{2})dy = 0

**Solution:**

We have,

(1 + y

^{2})tan^{-1}xdx + 2y(1 + x^{2})dy = 0 -(i)On integrating both sides,

-(ii)

Let, I =

2I = (1/2)(tan

^{-1}x)^{2}I = (1/4)(tan

^{-1}x)^{2}From equation (ii)

(1/2)log(1 + y

^{2}) = -(1/4)(tan^{-1}x)^{2 }+ clog(1 + y

^{2}) + (1/2)(tan^{-1}x)^{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}