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

### 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) = (ex + 1)y

Solution:

We have,

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

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

On integrating both sides

∫(dy/y) = ∫(ex + 1)dx

log(y) = (ex + 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) = exy4

Solution:

We have,

5(dy/dx) = exy4

5(dy/y4) = ex

On integrating both sides,

5∫(dy/y4) = ∫ex

-(5/3)(1/y3) = ex + 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) = exey + x2ey

dy = ey(ex + x2)dx

e-ydy = (ex + x2)dx

On integrating both sides,

∫e-ydy = ∫(ex + x2)dx

-e-y = ex + (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. (ey + 1)cosxdx + eysinxdy = 0

Solution:

We have,

(ey + 1)cosxdx + eysinxdy = 0

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

On integrating both sides,

∫(cosx/sinx)dx = -∫[ey/(ey + 1)]dy

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

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

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

sinx(ey + 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) = exey + x3ey

dy = ey(ex + x3)dx

e-ydy = (ex + x3)dx

On integrating both sides,

∫e-ydy = ∫(ex + x3)dx

-e-y = ex + (x4/4) + c

e-y + ex + (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 + y2 = z2

On differentiating both sides

2ydy = 2zdz

ydy = zdz  = ∫[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 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 + x2 = z2

On differentiating both sides

2xdx = 2zdz

xdx = zdz  = -∫[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 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)  (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)

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