# RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations

Here we provide RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 Differential Equations book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 22.7 Solutions Chapter 22 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,

(x+ 1)dy = xydx

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

On integrating both sides

∫(dy/y) = ∫[x/(x+ 1)]dx

log(y) = (1/2)∫[2x/(x+ 1)]dx

log(y) = (1/2)log(x+ 1) + c (Where ‘c’ is integration constant)

### Question 3. (dy/dx) = (ex + 1)y

Solution:

We have,

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

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

On integrating both sides

∫(dy/y) = ∫(e+ 1)dx

log(y) = (e+ 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∫[x+ x + 1 + 1/(x – 1)]dx

log(y) = (2/3)(x3) + x+ 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 = (x+ 1)dx

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

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

On integrating both sides,

∫(y+ 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) = e+ 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) = exe+ x2ey

dy = ey(e+ x2)dx

e-ydy = (e+ x2)dx

On integrating both sides,

∫e-ydy = ∫(e+ x2)dx

-e-y = e+ (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) = y– y

[1/(y– y)]dy = dx/x

On integrating both sides,

∫[1/(y– 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,

(e+ 1)cosxdx + eysinxdy = 0

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

On integrating both sides,

∫(cosx/sinx)dx = -∫[ey/(e+ 1)]dy

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

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

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

sinx(e+ 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

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) = exe+ x3ey

dy = ey(e+ x3)dx

e-ydy = (e+ x3)dx

On integrating both sides,

∫e-ydy = ∫(e+ x3)dx

-e-y = e+ (x4/4) + c

e-y + e+ (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

On integrating both sides,

Let, 1 + y= z2

On differentiating both sides

2ydy = 2zdz

ydy = zdz  = ∫[z2/(z– 1)]dz

= ∫[1 + 1/(z– 1)]dz

= z + (1/2)log[(z – 1)/(z + 1)]

On putting the value of z in above equation Similarly, =  (Where ‘c’ is integration constant)

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

Solution:

We have,

√(1 + x2)(dy) + √(1 + y2)dx = 0

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. Solution:

We have,

On integrating both sides,

Let, 1 + x= z2

On differentiating both sides

2xdx = 2zdz

xdx = zdz  = -∫[z2/(z– 1)]dz

= -∫[1 + 1/(z– 1)]dz

= -z – (1/2)log[(z – 1)/(z + 1)]

On putting the value of z in above equation

Let, 1 + y= v2

On differentiating both sides

2ydy = 2vdv

ydy = vdv

= ∫(vdv/v)

= v

On putting the value of v in above equation

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

### Question 19. Solution:

We have,

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

On integrating both sides,

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

Since, ∫ex(sin2x + sin2x)dx = exsin2

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∫{ ∫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)

### 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

On integrating both sides,

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

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

On integrating both sides,

√(1 – y2) = -√(1 – x2) + c

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

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

Solution:

We have,

y(1 + ex)dy =(y + 1)exdx

On integrating both sides,

∫[1 – 1/(y + 1)]dy = ∫exdx/(1 + ex)

y – log(y + 1) = log(1 + ex) + 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,

(y+ 1)dx – (x+ 1)dy = 0

(y+ 1)dx = (x+ 1)dy

On integrating both sides,

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)

On integrating both sides,

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

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

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

Solution:

We have,

(dy/dx) = ex+y + e-x+y

(dy/dx) = ex.e+ e-x.ey

(dy/dx) = ey(e+ e-x)

dy/e= (e+ e-x)dx

On integrating both sides,

∫e-ydy = ∫exdx + ∫e-xdx

-e-y = e– e-x + c

e-x-e-y = e+ 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,

(xy+ 2x)(dx) + (x2y + 2y)dy = 0

x(y+ 2)(dx) = -y(x+ 2)dy

Multiplying both sides by 2,

On integrating both sides,

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

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

Solution:

We have,

cosecx logy(dy/dx) + x2y= 0

log(y)dy/y= -x2dx/cosecx

On integrating both sides,

∫[log(y)/y2]dy = -∫x2sinxdx

-log(y)/y + ∫dy/y= 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)

On integrating both sides,

Multiplying both sides by 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). yex/ydx = (xex/y + y2)dy

Solution:

We have,

yex/ydx = (xex/y + y2)dy

yex/ydx – xex/ydy = y2dy

ex/y(ydx – xdy)/y= dy

ex/yd(x/y) = dy

On integrating both sides,

∫ex/yd(x/y) = ∫dy

ex/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)

On integrating both sides, -(ii)

Let, 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)+ c

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

### 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)

y= x3c

Put x = 1, y = 2 in above equation

c = 4

y= 4x

### 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) = 2exy3, y(0) = 1/2

Solution:

We have,

(dy/dx) = 2exy3

dy/y= 2exdx

On integrating both sides,

∫dy/y= 2∫exdx

-(1/2y2) = 2e+ c

Put x = 0, y = (1/2) in above equation

-(4/2) = 2 + c

c = -4

-(1/2y2) = 2e– 4

y2(4e– 8) = -1

y2(8 – 4ex) = 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) = r = r0 ### 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 = ### 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)

y= x5

y = |x|(5/2)

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

Solution:

We have,

(dy/dx) = 2e2xy

(dy/y2) = 2e2xdx

On integrating both sides

∫(dy/y2) = 2∫e2xdx

-(1/y) = 2e2x/2 + c

-(1/y) = e2x + c

Put x = 0, y = -1 in above equation

1 = e+ c

c = 0

-(1/y) = e2x

y = -e-2x

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

Solution:

We have,

cosy(dy/dx) = e

cosydy = exdx

On integrating both sides

∫cosydy = ∫exdx

siny = e+ c

Put x = 1, y = π/2 in above equation

sin(π/2) = e+ c

1 = 1 + c

c = 0

siny = ex

y = sin-1(ex)

### 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) = x+ 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 + x+ y+ x2y2

(dy/dx) = (1 + x2) + y2(1 + x2)

(dy/dx) = (1 + x2)(1 + y2)

On integrating both sides

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 + y+ xy

(dy/dx) = (1 + x) + y2(1 + x)

(dy/dx) = (1 + x)(1 + y2)

On integrating both sides

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)

On integrating both sides

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 = (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∫[( ∫xdx]dx + x+ c

-cosy + ysiny – ∫sinydy = x2logx – ∫xdx + x+ c

-cosy + ysiny + cosy = x2logx – x2/2 + x+ 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 + 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.

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 – ∫[ ∫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) = -4xy

(dy/y2) + 4xdx = 0

On integrating both sides

∫(dy/y2) + 4∫xdx = 0

-(1/y) + 2x= c

Put x = 0, y = 1 in above equation

-1 + 0 = c

c = -1

-(1/y) + 2x= -1

(1/y) = 2x+ 1

y = 1/(2x+ 1)

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

Solution:

We have,

(dy/dx) = exsinx

dy = exsinxdx

On integrating both sides

∫dy = ∫exsinxdx

Let, I = ∫exsinxdx

I = ex∫sinx – ∫[ ∫sinxdx]dx

I = -excosx + ∫excosxdx

I = -excosx + ex∫cosxdx – ∫[ ∫cosxdx]dx

I = -excosx + exsinx – ∫exsinxdx

I = -excosx + exsinx – I

2I = -excosx + exsinx

I = ex(sinx – cosx)/2

y = ex(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)+ 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

(4/3)π.3r2(dr/dt) = k

4πr2dr = kdt

On integrating both sides

∫4πr2dr = ∫kdt

4π(r3/3) = kt + c

4πr= 3(kt + c)           -(i)

At t = 0, r = 38

4π(3)= 3(0 + c)

c = 36π

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

4π(6)= 3(kt + 36π)

864π = 9k + 108π

k = 84π

4πr= 3(84πt + 36π)

r= 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 = 1000e0.5

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

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

t t1 ### Question 58. If y(x) is a solution of the differential equation , and y(0) = 1, then find the value of y(π/2).

Solution:

We have, (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)

### 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 = v+ 1

⇒ 1/(v+ 1) dv = dx

Integrating both sides, we get

∫ 1/(v+ 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|

⇒ ey – C1 = x + y + 2

⇒ e/ eC1 = x + y + 2

⇒ e– C ey = x + y + 2

⇒ C ey  = x + y + 2                  (C = e– C1)

⇒ x =  C ey – y – 2

### Question 11: dy/dx + 1 = ex + y

Solution:

We have,

dy/dx + 1 = ex + 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 = et

⇒ e– t dt = dx

⇒ – e– t = x + C

⇒ – e– (x + y) = x + C         [Since t = x + y]

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.