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

### Solve the following 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,

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. (x^{2 }+ 1)dy = xydx

**Solution:**

We have,

(x

^{2 }+ 1)dy = xydx(dy/y) = [x/(x

^{2 }+ 1)]dxOn integrating both sides

âˆ«(dy/y) = âˆ«[x/(x

^{2 }+ 1)]dxlog(y) = (1/2)âˆ«[2x/(x

^{2 }+ 1)]dxlog(y) = (1/2)log(x

^{2 }+ 1) + c (Where ‘c’ is integration constant)

### Question 3. (dy/dx) = (e^{x }+ 1)y

**Solution:**

We have,

(dy/dx) = (e

^{x }+ 1)y(dy/y) = (e

^{x }+ 1)dxOn integrating both sides

âˆ«(dy/y) = âˆ«(e

^{x }+ 1)dxlog(y) = (e

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

### Question 4. (x – 1)(dy/dx) = 2x^{3}y

**Solution:**

We have,

(x – 1)(dy/dx) = 2x

^{3}y(dy/y) = [2x

^{3}/(x – 1)]dxOn integrating both sides

âˆ«(dy/y) = âˆ«[2x

^{3}/(x – 1)]dxâˆ«(dy/y) = 2âˆ«[x

^{2 }+ x + 1 + 1/(x – 1)]dxlog(y) = (2/3)(x

^{3}) + x^{2 }+ 2x + 2log(x – 1) + c (Where ‘c’ is integration constant)

### Question 5. xy(y + 1)dy = (x^{2 }+ 1)dx

**Solution:**

We have,

xy(y + 1)dy = (x

^{2 }+ 1)dxy(y + 1)dy = [(x

^{2 }+ 1)/x]dx(y

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

âˆ«(y

^{2 }+ y)dy = âˆ«xdx + (dx/x)(y

^{3}/3) + (y^{2}/2) = (x^{2}/2) + log(x) + c (Where ‘c’ is integration constant)

### Question 6. 5(dy/dx) = e^{x}y^{4}

**Solution:**

We have,

5(dy/dx) = e

^{x}y^{4}5(dy/y

^{4}) = e^{x}On integrating both sides,

5âˆ«(dy/y

^{4}) = âˆ«e^{x}-(5/3)(1/y

^{3}) = e^{x }+ c (Where ‘c’ is integration constant)

### Question 7. xcosydy = (xe^{x}logx + e^{x})dx

**Solution:**

We have,

xcosydy = (xe

^{x}logx + e^{x})dxcosydy = e

^{x}(logx + 1/x)dxOn integrating both sides,

âˆ«cosydy = âˆ«e

^{x}(logx + 1/x)dxSince, âˆ«[f(x) + f'(x)]e

^{x}dx] = e^{x}f(x)siny = e

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

### Question 8. (dy/dx) = e^{x+y }+ x^{2}e^{y}

**Solution:**

We have,

(dy/dx) = e

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

^{x}e^{y }+ x^{2}e^{y}dy = e

^{y}(e^{x }+ x^{2})dxe

^{-y}dy = (e^{x }+ x^{2})dxOn integrating both sides,

âˆ«e

^{-y}dy = âˆ«(e^{x }+ x^{2})dx-e

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

### Question 9. x(dy/dx) + y = y^{2}

**Solution:**

We have,

x(dy/dx) + y = y

^{2}x(dy/dx) = y

^{2 }– y[1/(y

^{2 }– y)]dy = dx/xOn integrating both sides,

âˆ«[1/(y

^{2 }– y)]dy = âˆ«dx/xâˆ«[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. (e^{y }+ 1)cosxdx + e^{y}sinxdy = 0

**Solution:**

We have,

(e

^{y }+ 1)cosxdx + e^{y}sinxdy = 0(cosx/sinx)dx = -[e

^{y}/(e^{y }+ 1)]dyOn integrating both sides,

âˆ«(cosx/sinx)dx = -âˆ«[e

^{y}/(e^{y }+ 1)]dylog(sinx) = -log(e

^{y }+ 1) + log(c)log(sinx) + log(e

^{y }+ 1) = log(c)log[sinx(e

^{y }+ 1)] = log(c)sinx(e

^{y }+ 1) = c (Where ‘c’ is integration constant)

### Question 11. xcos^{2}ydx = ycos^{2}xdy

**Solution:**

We have,

xcos

^{2}ydx = ycos^{2}xdy(x/cos

^{2}x)dx = (y/cos^{2}y)dyxsec

^{2}xdx = ysec^{2}ydyOn integrating both sides,

âˆ«xsec

^{2}xdx = âˆ«ysec^{2}ydyxtanx – 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) = (xe^{x}logx + e^{x})/(xcosy)

**Solution:**

We have,

(dy/dx) = (xe

^{x}logx + e^{x})/(xcosy)xcosydy = (xe

^{x}logx + e^{x})dxcosydy = e

^{x}(logx + 1/x)dxOn integrating both sides,

âˆ«cosydy = âˆ«e

^{x}(logx + 1/x)dxSince, âˆ«[f(x) + f'(x)]e

^{x}dx] = e^{x}f(x)siny = e

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

### Question 15. (dy/dx) = e^{x+y }+ x^{3}e^{y}

**Solution:**

We have,

(dy/dx) = e

^{x+y }+ x^{3}e^{y}(dy/dx) = e

^{x}e^{y }+ x^{3}e^{y}dy = e

^{y}(e^{x }+ x^{3})dxe

^{-y}dy = (e^{x }+ x^{3})dxOn integrating both sides,

âˆ«e

^{-y}dy = âˆ«(e^{x }+ x^{3})dx-e

^{-y }= e^{x }+ (x^{4}/4) + ce

^{-y }+ e^{x }+ (x^{4}/4) = c (Where ‘c’ is integration constant)

### Question 16. yâˆš(1 + x^{2}) + xâˆš(1 + y^{2})(dy/dx) = 0

**Solution:**

We have,

yâˆš(1 + x

^{2}) + âˆš(1 + y^{2})(dy/dx) = 0yâˆš(1 + x

^{2})dx = -xâˆš(1 + y^{2})dyOn integrating both sides,

Let, 1 + y

^{2 }= z^{2}On differentiating both sides

2ydy = 2zdz

ydy = zdz

=

=

= âˆ«[z

^{2}/(z^{2 }– 1)]dz= âˆ«[1 + 1/(z

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

**Solution:**

We have,

âˆš(1 + x

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

log[y + âˆš(1 + y^{2})] = -log[x + âˆš(1 + x^{2})] + logclog[y + âˆš(1 + y

^{2})] + log[x + âˆš(1 + x^{2})] = logclog([y + âˆš(1 + y

^{2})][x + âˆš(1 + x^{2})]) = logc[y + âˆš(1 + y

^{2})][x + âˆš(1 + x^{2})] = c (Where ‘c’ is integration constant)

### Question 18.

**Solution:**

We have,

On integrating both sides,

Let, 1 + x

^{2 }= z^{2}On differentiating both sides

2xdx = 2zdz

xdx = zdz

=

=

= -âˆ«[z

^{2}/(z^{2 }– 1)]dz= -âˆ«[1 + 1/(z

^{2 }– 1)]dz= -z – (1/2)log[(z – 1)/(z + 1)]

On putting the value of z in above equation

Let, 1 + y

^{2 }= v^{2}On differentiating both sides

2ydy = 2vdv

ydy = vdv

= âˆ«(vdv/v)

= v

On putting the value of v in above equation

= âˆš(1 + y

^{2})=

= (Where ‘c’ is integration constant)

### Question 19.

**Solution:**

We have,

y(2logy + 1)dy = e

^{x}(sin^{2}x + sin2x)dxOn integrating both sides,

âˆ«y(2logy + 1)dy = âˆ«e

^{x}(sin^{2}x + sin2x)dxSince, âˆ«e

^{x}(sin^{2}x + sin2x)dx = e^{x}sin^{2}xUsing property âˆ«[f(x) + f'(x)]e

^{x}= e^{x}f(x)y

^{2}log(y) – âˆ«ydy + y^{2}/2 = e^{x}sin^{2}x + cy

^{2}log(y) – y^{2}/2 + y^{2}/2 = e^{x}sin^{2}x + cy

^{2}log(y) = e^{x}sin^{2}x + 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 = x

^{2}logx – âˆ«xdx + (x^{2}/2) + c-cosy + ysiny + cosy = x

^{2}logx – (x^{2}/2) + (x^{2}/2) + cysiny = x

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

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