# 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,

âˆ«(dy/y) = âˆ«[2x + (x – 1)]dx

log(y) = âˆ«[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

âˆ«(dy/y) = âˆ«[x/(x2 + 1)]dx

log(y) = (1/2)âˆ«[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

âˆ«(dy/y) = âˆ«(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

âˆ«(dy/y) = âˆ«[2x3/(x – 1)]dx

âˆ«(dy/y) = 2âˆ«[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,

âˆ«(y2 + y)dy = âˆ«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,

5âˆ«(dy/y4) = âˆ«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,

âˆ«cosydy = âˆ«ex(logx + 1/x)dx

Since, âˆ«[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,

âˆ«e-ydy = âˆ«(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,

âˆ«[1/(y2 – y)]dy = âˆ«dx/x

âˆ«[1/(y – 1) – 1/y]dy = âˆ«(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,

âˆ«(cosx/sinx)dx = -âˆ«[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,

âˆ«xsec2xdx = âˆ«ysec2ydy

xtanx – âˆ«tanxdx = ytany – âˆ«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,

âˆ«[y/(y – 1)]dy = âˆ«[(x + 1)/x]dx

âˆ«[1 + 1/(y – 1)]dy = âˆ«[(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,

âˆ«tanydy = -âˆ«(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,

âˆ«cosydy = âˆ«ex(logx + 1/x)dx

Since, âˆ«[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,

âˆ«e-ydy = âˆ«(ex + x3)dx

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

e-y + ex + (x4/4) = c  (Where ‘c’ is integration constant)

### Question 16. yâˆš(1 + x2) + xâˆš(1 + y2)(dy/dx) = 0

Solution:

We have,

yâˆš(1 + x2) + âˆš(1 + y2)(dy/dx) = 0

yâˆš(1 + x2)dx = -xâˆš(1 + y2)dy

On integrating both sides,

Let, 1 + y2 = z2

On differentiating both sides

2ydy = 2zdz

ydy = zdz

= âˆ«[z2/(z2 – 1)]dz

= âˆ«[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. âˆš(1 + x2)(dy) + âˆš(1 + y2)dx = 0

Solution:

We have,

âˆš(1 + x2)(dy) + âˆš(1 + y2)dx = 0

On integrating both sides,
log[y + âˆš(1 + y2)] = -log[x + âˆš(1 + x2)] + logc

log[y + âˆš(1 + y2)] + log[x + âˆš(1 + x2)] = logc

log([y + âˆš(1 + y2)][x + âˆš(1 + x2)]) = logc

[y + âˆš(1 + y2)][x + âˆš(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

= -âˆ«[z2/(z2 – 1)]dz

= -âˆ«[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

= âˆ«(vdv/v)

= v

On putting the value of v in above equation

= âˆš(1 + y2)

(Where ‘c’ is integration constant)

### Question 19.

Solution:

We have,

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

On integrating both sides,

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

Since, âˆ«ex(sin2x + sin2x)dx = exsin2

Using property âˆ«[f(x) + f'(x)]ex = exf(x)

y2log(y) – âˆ«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,

âˆ«(siny + ycosy)dy = âˆ«x(2logx + 1)dx

âˆ«sinydy + yâˆ«cosydy – âˆ«{(dy/dy)âˆ«cosydy}dy = 2logxâˆ«xdx – 2âˆ«{âˆ«xdx} + âˆ«xdx

-cosy + ysiny – âˆ«sinydy = x2logx – âˆ«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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