# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.9 | Set 3

### Question 27. (x^{2 }– 2xy)dy + (x^{2 }– 3xy + 2y^{2})dx = 0

**Solution:**

We have,

(x

^{2 }– 2xy)dy + (x^{2 }– 3xy + 2y^{2})dx = 0(dy/dx) = (x

^{2 }– 3xy + 2y^{2})/(2xy – y^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }– 3xvx + 2v^{2}x^{2})/(2xvx – x^{2})v + x(dv/dx) = (1 – 3v + 2v

^{2})/(2v – 1)x(dv/dx) = [(1 – 3v + 2v

^{2})/(2v – 1)] – vx(dv/dx) = (1 – 3v + 2v

^{2 }– 2v^{2 }+ v)/(2v – 1)x(dv/dx) = (1 – 2v)/(2v – 1)

x(dv/dx) = -1

dv = -(dx/x)

On integrating both sides,

âˆ«dv = -âˆ«(dx/x)

v = -log|x| + log|c|

(y/x) + log|x| = log|c| (Where â€˜câ€™ is integration constant)

### Question 28. x(dy/dx) = y – xcos^{2}(y/x)

**Solution:**

We have,

x(dy/dx) = y – xcos

^{2}(y/x)(dy/dx) = y/x – cos

^{2}(y/x)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – cos

^{2}(v)x(dv/dx) = -cos

^{2}(v)dv/cos

^{2}(v) = -(dx/x)On integrating both sides,

âˆ«dv/cos

^{2}(v) = -âˆ«(dx/x)âˆ«sec

^{2}vdv = -âˆ«(dx/x)tan(v) = -log|x| + log|c|

tan(y/x) = log|c/x| (Where â€˜câ€™ is integration constant)

### Question 29. x(dy/dx) – y = 2âˆš(y^{2 }– x^{2})

**Solution:**

We have,

x(dy/dx) – y = 2âˆš(y

^{2 }– x^{2})(dy/dx) = [2âˆš(y

^{2 }– x^{2}) + y]/xIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

x(dv/dx) = 2âˆš(v

^{2 }– 1)dv/âˆš(v

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

âˆ«dv/âˆš(v

^{2 }– 1) = 2âˆ«(dx/x)log|v + âˆš(v

^{2}– 1)| = 2log(x) + log(c)|v + âˆš(v

^{2 }– 1)| = |cx^{2}|(Where â€˜câ€™ is integration constant)

### Question 30. xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

**Solution:**

We have,

xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

xycos(y/x)dx + x

^{2}cos(y/x)dy = xysin(y/x)dy – y^{2}sin(y/x)dxx

^{2}cos(y/x)dy – xysin(y/x)dy = -y^{2}sin(y/x)dx – xycos(y/x)dxIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vcosv + v

^{2}sinv)/(vsinv – cosv)x(dv/dx) = [(vcosv + v

^{2}sinv)/(vsinv – cosv)] – vx(dv/dx) = (vcosv + v

^{2}sinv – v^{2}sinv + vcosv)/(vsinv – cosv)x(dv/dx) = (2vcosv)/(vsinv – cosv)

[(vsinv – cosv)/(vcosv)]dv = 2(dx/x)

On integrating both sides,

âˆ«tanvdv – âˆ«(dv/v) = 2log|x| + log|c|

log|secv| – log|v| = log|cx

^{2}|log|(secv/v)| = log|cx

^{2}|(x/y)sec(y/x) = cx

^{2}sec(y/x) = cxy (Where â€˜câ€™ is integration constant)

### Question 31. (x^{2 }+ 3xy + y^{2})dx – x^{2}dy = 0

**Solution:**

We have,

(x

^{2 }+ 3xy + y^{2})dx – x^{2}dy = 0dy/dx = (x

^{2 }+ 3xy + y^{2})/x^{2}It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }+ 3xvx + v^{2}x^{2})/x^{2}v + x(dv/dx) = (1 + 3v + v

^{2})x(dv/dx) = (1 + 3v + v

^{2}) – vx(dv/dx) = (1 + 2v + v

^{2})x(dv/dx) = (1 + v)

^{2}dv/(1 + v)

^{2 }= (dx/x)On integrating both sides,

âˆ«dv/(1 + v)

^{2 }= âˆ«(dx/x)-[1/(v + 1)] = log|x| – c

x/(x + y) + log|x| = c (Where â€˜câ€™ is an integration constant)

### Question 32. (x – y)(dy/dx) = (x + 2y)

**Solution:**

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(x – y)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + 2vx)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v

^{2})/(1 – v)x(dv/dx) = (1 + v + v

^{2})/(1 – v)(1 – v)dv/(1 + v + v

^{2}) = (dx/x)On integrating both sides,

âˆ«[(1 – v)/(1 + v + v

^{2})]dv = âˆ«(dx/x)(Where â€˜câ€™ is an integration constant)

### Question 33. (2x^{2}y + y^{3})dx + (xy^{2 }– 3x^{2})dy = 0

**Solution:**

We have,

(2x

^{2}y + y^{3})dx + (xy^{2 }– 3x^{2})dy = 0dy/dx = (2x

^{2}y + y^{3})/(3x^{3 }– xy^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (2x

^{2}vx + v^{3}x^{3})/(3x^{3 }– xv^{2}x^{2})v + x(dv/dx) = (2v + v

^{3})/(3 – v^{3})x(dv/dx) = [(2v + v

^{3})/(3 – v^{3})] – vx(dv/dx) = (2v + v

^{3 }– 3v + v^{3})/(3 – v^{3})(3 – v

^{3})dv/(2v^{3 }– v) = (dx/x)On integrating both sides,

âˆ«[(3 – v

^{3})/(2v^{3 }– v)]dv = âˆ«(dx/x)Using partial fraction,

3 – v

^{2 }= A(2v^{2 }– 1) + (Bv + C)v3 – v

^{2 }= 2Av^{2 }– A + Bv^{2 }+ Cv3 – v

^{2 }= v^{2}(2A + B) + Cv – AOn comparing the coefficients, we get

A = -3,

B = 5,

C = 0,

-3log|v|+(5/4)log|2v

^{2}-1|=log|x|+log|c|-12log|v|+5log|2v

^{2}-1|=4log|x|+4log|c||2y

^{2 }– x^{2}|^{5 }= x^{2}c^{4}y^{12}(Where â€˜câ€™ is an integration constant)

### Question 34. x(dy/dx) – y + xsin(y/x) = 0

**Solution:**

We have,

x(dy/dx) – y + xsin(y/x) = 0

x(dy/dx) = y – xsin(y/x)(dy/dx) = [y – xsin(y/x)]/x

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = [vx – xsinv]/x

v + x(dv/dx) = (v – sinv)

x(dv/dx) = -sinv

cosecvdv = -(dx/x)

On integrating both sides,

âˆ«cosecvdv = -âˆ«(dx/x)

-log|cosecv + cotv| = -log|x| + log|c|

-log|(1/sinv) + (cosv/sinv)| = -log|x/c|

|(1 + cosv)/sinv| = |x/c|

xsinv = c(1 + cosv)

xsin(y/x) = c[1 + cos(y/x)] (Where â€˜câ€™ is integration constant)

### Question 35. ydx + {xlog(y/x)}dy – 2xydy = 0

**Solution:**

We have,

ydx + {xlog(y/x)}dy – 2xydy = 0

y + {xlog(y/x)}(dy/dx) – 2xy = 0

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v/(2 – logv)

x(dv/dx) = [v/(2 – logv)] – v

x(dv/dx) = (v – 2v + vlogv)/(2 – logv)

x(dv/dx) = -v(logv – 1)/(logv – 2)

On integrating both sides,

Let. logv – 1 = z

On differentiating both sides,

(dv/v) = dz

âˆ«dz – âˆ«(dz/z) = -âˆ«(dx/x)

z – log|z| = -log|x| + log|c|

(logv – 1) – log|(logv – 1)| = -log|x| + log|c|

logv – log|logv – 1| = -log|x| + log|c| + 1

log|(logv – 1)/v| = log|c

_{1}x||logv – 1| = |c

_{1}xv||log(y/x) – 1| = |c

_{1}x(y/x)||log(y/x) – 1| = |c

_{1}y| (Where â€˜c_{1}â€™ is integration constant)

### Question 36(i). (x^{2 }+ y^{2})dx = 2xydy, y(1) = 0

**Solution:**

We have,

(x

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

^{2 }+ y^{2})/2xyIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x

^{2 }+ v^{2}x^{2})/2vx^{2}v + x(dv/dx) = (1 + v

^{2})/2vx(dv/dx) = [(1 + v

^{2})/2v] – vx(dv/dx) = (1 + v

^{2 }– 2v^{2})/2vx(dv/dx) = (1 – v

^{2})/2v2vdv/(1 – v

^{2}) = (dx/x)On integrating both sides,

âˆ«2vdv/(1 – v

^{2}) = âˆ«(dx/x)-log|1 – v

^{2}| = log|x| – log|c|log|1 – v

^{2}| = log|c/x||1 – y

^{2}/x^{2}| = |c/x||x

^{2 }– y^{2}| = |cx|At x = 1, y = 0

1 – 0 = c

c = 1

|x

^{2 }– y^{2}| = |x|(x

^{2 }– y^{2})^{ }= x

### Question 36(ii). xe^{x/y }– y + x(dy/dx) = 0, y(e) = 0

**Solution:**

We have,

xe

^{x/y }– y + x(dy/dx) = 0(dy/dx) = (y – xe

^{x/y})/xIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vx – xe

^{v})/xv + x(dv/dx) = v – e

^{v}x(dv/dx) = v – e

^{v }– vx(dv/dx) = -e

^{v}e

^{-v}dv = -(dx/x)On integrating both sides,

âˆ«e

^{-v}dv = -âˆ«(dx/x)-e

^{-v }= -log|x| – log|c|e

^{-v }= log|x| + log|c|e

^{-(y/x) }= log|x| + log|c|At x = e, y = 0

e

^{-(0/e) }= log|e| + log|c|1 = 1 + log|c|

c = 0

e

^{-y/x }= logx

### Question 36(iii). (dy/dx) – (y/x) + cosec(y/x) = 0, y(1) = 0

**Solution:**

We have,

(dy/dx) – (y/x) + cosec(y/x) = 0

(dy/dx) = (y/x) – cosec(y/x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – cosec(v)

x(dv/dx) = v – cosec(v) – v

x(dv/dx) = -cosec(v)

-sin(v)dv = (dx/x)

On integrating both sides,

-âˆ«sin(v)dv = âˆ«(dx/x)

cos(v) = log|x| + log|c|

cos(y/x) = log|x| + log|c|

At x = 1, y = 0

cos(0/1) = log|1| + log|c|

1 = 0 + log|c|

log|c| = 1

cos(y/x) = log|x| + 1

log|x| = cos(y/x) – 1

### Question 36(iv). (xy – y^{2})dx – x^{2}dy = 0, y(1) = 1

**Solution:**

We have,

(xy – y

^{2})dx – x^{2}dy = 0(dy/dx) = (xy – y

^{2})/x^{2}(dy/dx) = (y/x) – (y

^{2}/x^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – v

^{2}x(dv/dx) = v – v

^{2 }– vx(dv/dx) = -v

^{2}-(dv/v

^{2}) = (dx/x)On integrating both sides,

-âˆ«(dv/v

^{2}) = âˆ«(dx/x)-(-1/v) = log|x| + c

(1/v) = log|x| + c

x/y = log|x| + c

At x = 1, y = 1

1 = log|1| + c

c = 1

x/y = log|x| + 1

y = x/[log|x| + 1]

### Question 36(v). (dy/dx) = [y(x + 2y)]/[x(2x + y)]

**Solution:**

We have,

(dy/dx) = [y(x + 2y)]/[x(2x + y)]

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = [vx(x + 2vx)]/[x(2x + vx]

x(dv/dx) = [vx(x + 2vx)]/[x(2x + vx] – v

x(dv/dx) = (v + 2v

^{2 }– 2v – v^{2})/(2 + v)x(dv/dx) = (v

^{2 }– v)/(2 + v)(2 + v)dv/[v(v – 1)] = (dx/x)

On integrating both sides,

Using partial derivative,

2 + v = A(v – 1) + B(v)

2 + v = v(A + B) – A

On comparing the coefficients,

A = -2

B = 3

-2âˆ«(dv/v) + 3âˆ«dv/(v – 1) = âˆ«(dx/x)

-2log|v| + 3log|v – 1| = log|x| + log|c|

log|(v – 1)

^{3}/v^{2}| = log|xc|(v – 1)

^{3 }= v^{2}|xc|(y – x)

^{3}/x^{3 }= (y/x)^{2}|xc|(y – x)

^{3}= y^{2}x^{2}cAt x = 1, y = 2,

(2 – 1)

^{3}= 4 * 1 * cc = (1/4)

(y – x)

^{3}= (1/4)y^{2}x^{2}

### Question 36(vi). (y^{4 }– 2x^{3}y)dx + (x^{4 }– 2xy^{3})dy = 0, y(1) = 0

**Solution:**

We have,

(y

^{4 }– 2x^{3}y)dx + (x^{4 }– 2xy^{3})dy = 0dy/dx = (2x

^{3}y – y^{4})/(x^{4 }– 2xy^{3})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (2x

^{3}vx – v^{4}x^{4})/(x^{4 }– 2xv^{3}x^{3})v + x(dv/dx) = (2v – v

^{4})/(1 – 2v^{3})x(dv/dx) = [(2v – v

^{4})/(1 – 2v^{3})] – vx(dv/dx) = (2v – v

^{4 }– v + 2v^{4})/(1 – 2v^{3})x(dv/dx) = (v + v

^{4})/(1 – 2v^{3})On integrating both sides,

âˆ«(dv/v) – âˆ«(3v

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

^{3}| = log|xc|log|v/(1 + v

^{3})| = log|xc|At x = 1, y = 1,

1/(1 + 1) = c

c = (1/2)

(yx

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

### Question 36(vii). x(x^{2 }+ 3y^{2})dx + y(y^{2 }+ 3x^{2})dy = 0, y(1) = 1

**Solution:**

We have,

x(x

^{2 }+ 3y^{2})dx + y(y^{2 }+ 3x^{2})dy = 0dy/dx = -[x(x

^{2 }+ 3y^{2})/y(y^{2 }+ 3x^{2})]It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

x(dv/dx) = -(1 + 3v

^{2 }+ v^{4 }+ 3v^{2})/v(v^{2 }+ 3)[(v

^{3 }+ 3v)/(1 + 6v^{2 }+ v^{4})]dv = -(dx/x)Multiply both sides with 4 and integrating,

log|v

^{4 }+ 6v^{2 }+ 1| = -log|x|^{4 }+ log|c||v

^{4 }+ 6v^{2 }+ 1| = |c/x^{4}|(y

^{4 }+ 6x^{2}y^{2 }+ x^{4}) = cAt y = 1, x = 1

(1 + 6 + 1) = c

c = 8

(y

^{4 }+ 6x^{2}y^{2 }+ x^{4}) = 8

### Question 36(viii). {xsin^{2}(y/x) – y}dx + xdy = 0, y(1) = Ï€/4

**Solution:**

We have,

{xsin

^{2}(y/x) – y}dx + xdy = 0dy/dx = [y – xsin

^{2}(y/x)]/xdy/dx = (y/x) – sin

^{2}(y/x)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – sin

^{2}(v)x(dv/dx) = v – sin

^{2}(v) – vx(dv/dx) = -sin

^{2}(v)-cosec

^{2}(v)dv = (dx/x)On integrating both sides,

-âˆ«cosec

^{2}(v) = âˆ«(dx/x)cot(v) = log|x| + log|c|

cot(y/x) = log|xc|

At x = 1, y = Ï€/4

cot(Ï€/4) = log|c|

log|c| = 1

c = e

cot(y/x) = log|ex|

### Question 36(ix). x(dy/dx) – y + xsin(y/x) = 0, y(2) = Ï€

**Solution:**

We have,

x(dy/dx) – y + xsin(y/x) = 0

x(dy/dx) = y – xsin(y/x)

(dy/dx) = (y/x) – sin(y/x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – sin(v)

x(dv/dx) = v – sin(v) – v

x(dv/dx) = -sin(v)

-cosec(v)dv = (dx/x)

On integrating both sides,

âˆ«cosec(v) = -âˆ«(dx/x)

log|cosec(v) – cot(v)| = -log|x| + log|c|

log|cosec(v) – cot(v)| = -log|x| + log|c|

log|cosec(y/x) – cot(y/x)| = -log|x| + log|c|

At x = 2, y = Ï€

|cosec(Ï€/2) – cot(Ï€/2)| = -log|2| + log|c|

log|c| = 0

log|cosec(y/x) – cot(y/x)| = -log|x|

### Question 37. xcos(y/x)(dy/dx) = ycos(y/x) + x, When x = 1, y = Ï€/4

**Solution:**

We have,

xcos(y/x)(dy/dx) = ycos(y/x) + x

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

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v + 1/cosv

x(dv/dx) = v + 1/cosv – v

x(dv/dx) = 1/cosv

cosvdv = (dx/x)

On integrating both sides,

âˆ«cosvdv = âˆ«(dx/x)

sin(v) = log|x| + log|c|

sin(y/x) = log|x| + log|c|

At x = 1, y = Ï€/4

1/âˆš2 = 0 + log|c|

log|c| = (1/âˆš2)

sin(y/x) = log|x| + (1/âˆš2)

### Question 38. (x – y)(dy/dx) = (x + 2y), when x = 1,y = 0

**Solution:**

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(x – y)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + 2vx)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v

^{2})/(1 – v)(1 – v)dv/(1 + v + v

^{2}) = (dx/x)On integrating both sides,

At x = 1, y = 0

âˆš3tan

^{-1}|1/âˆš3| – (1/2)log|1| = cc = âˆš3(Ï€/6)

c = (Ï€/2âˆš3)

### Question 39. (dy/dx) = xy/(x^{2 }+ y^{2})

**Solution:**

We have,

(dy/dx) = xy/(x

^{2 }+ y^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = xvx/(x

^{2 }+ v^{2}x^{2})v + x(dv/dx) = v/(1 + v

^{2})x(dv/dx) = [v/(1 + v

^{2})] – vx(dv/dx) = (v – v – v

^{3})/(1 + v^{2})[-(1/v

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

-âˆ«dv/v

^{3 }– âˆ«dv/v = âˆ«(dx/x)(1/2v

^{2}) – log|v| = log|x| + c(x

^{2}/2y^{2}) = log|vx| + c(x

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

^{2}/2y^{2}) = log|y| + cAt x = 0, y = 1

c = 0

(x

^{2}/2y^{2}) = log|y|

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