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

• Last Updated : 18 Mar, 2021

### Question 27.  (x2 – 2xy)dy + (x2 – 3xy + 2y2)dx = 0

Solution:

We have,

(x2 – 2xy)dy + (x2 – 3xy + 2y2)dx = 0

(dy/dx) = (x2 – 3xy + 2y2)/(2xy – y2)

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) = (x2 – 3xvx + 2v2x2)/(2xvx – x2)

v + x(dv/dx) = (1 – 3v + 2v2)/(2v – 1)

x(dv/dx) = [(1 – 3v + 2v2)/(2v – 1)] – v

x(dv/dx) = (1 – 3v + 2v2 – 2v2 + 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 – xcos2(y/x)

Solution:

We have,

x(dy/dx) = y – xcos2(y/x)

(dy/dx) = y/x – cos2(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 – cos2(v)

x(dv/dx) = -cos2(v)

dv/cos2(v) = -(dx/x)

On integrating both sides,

âˆ«dv/cos2(v) = -âˆ«(dx/x)

âˆ«sec2vdv = -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(y2 – x2)

Solution:

We have,

x(dy/dx) – y = 2âˆš(y2 – x2)

(dy/dx) = [2âˆš(y2 – x2) + 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,

x(dv/dx) = 2âˆš(v2 – 1)

dv/âˆš(v2 – 1) = 2(dx/x)

On integrating both sides,

âˆ«dv/âˆš(v2 – 1) = 2âˆ«(dx/x)

log|v + âˆš(v2– 1)| = 2log(x) + log(c)

|v + âˆš(v2 – 1)| = |cx2|

(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 + x2cos(y/x)dy = xysin(y/x)dy – y2sin(y/x)dx

x2cos(y/x)dy – xysin(y/x)dy = -y2sin(y/x)dx – xycos(y/x)dx

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) = (vcosv + v2sinv)/(vsinv – cosv)

x(dv/dx) = [(vcosv + v2sinv)/(vsinv – cosv)] – v

x(dv/dx) = (vcosv + v2sinv – v2sinv + 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|cx2|

log|(secv/v)| = log|cx2|

(x/y)sec(y/x) = cx2

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

### Question 31. (x2 + 3xy + y2)dx – x2dy = 0

Solution:

We have,

(x2 + 3xy + y2)dx – x2dy = 0

dy/dx = (x2 + 3xy + y2)/x2

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) = (x2 + 3xvx + v2x2)/x2

v + x(dv/dx) = (1 + 3v + v2)

x(dv/dx) = (1 + 3v + v2) – v

x(dv/dx) = (1 + 2v + v2)

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 + v2)/(1 – v)

x(dv/dx) = (1 + v + v2)/(1 – v)

(1 – v)dv/(1 + v + v2) = (dx/x)

On integrating both sides,

âˆ«[(1 – v)/(1 + v + v2)]dv = âˆ«(dx/x)

(Where â€˜câ€™ is an integration constant)

### Question 33. (2x2y + y3)dx + (xy2 – 3x2)dy = 0

Solution:

We have,

(2x2y + y3)dx + (xy2 – 3x2)dy = 0

dy/dx = (2x2y + y3)/(3x3 – xy2)

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) = (2x2vx + v3x3)/(3x3 – xv2x2)

v + x(dv/dx) = (2v + v3)/(3 – v3)

x(dv/dx) = [(2v + v3)/(3 – v3)] – v

x(dv/dx) = (2v + v3 – 3v + v3)/(3 – v3)

(3 – v3)dv/(2v3 – v) = (dx/x)

On integrating both sides,

âˆ«[(3 – v3)/(2v3 – v)]dv = âˆ«(dx/x)

Using partial fraction,

3 – v2 = A(2v2 – 1) + (Bv + C)v

3 – v2 = 2Av2 – A + Bv2 + Cv

3 – v2 = v2(2A + B) + Cv – A

On comparing the coefficients, we get

A = -3,

B = 5,

C = 0,

-3log|v|+(5/4)log|2v2-1|=log|x|+log|c|

-12log|v|+5log|2v2-1|=4log|x|+4log|c|

|2y2 – x2|5 = x2c4y12  (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|c1x|

|logv – 1| = |c1xv|

|log(y/x) – 1| = |c1x(y/x)|

|log(y/x) – 1| = |c1y|  (Where â€˜c1â€™ is integration constant)

### Question 36(i). (x2 + y2)dx = 2xydy, y(1) = 0

Solution:

We have,

(x2 + y2)dx = 2xydy

(dy/dx) = (x2 + y2)/2xy

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) = (x2 + v2x2)/2vx2

v + x(dv/dx) = (1 + v2)/2v

x(dv/dx) = [(1 + v2)/2v] – v

x(dv/dx) = (1 + v2 – 2v2)/2v

x(dv/dx) = (1 – v2)/2v

2vdv/(1 – v2) = (dx/x)

On integrating both sides,

âˆ«2vdv/(1 – v2) = âˆ«(dx/x)

-log|1 – v2| = log|x| – log|c|

log|1 – v2| = log|c/x|

|1 – y2/x2| = |c/x|

|x2 – y2| = |cx|

At x = 1, y = 0

1 – 0 = c

c = 1

|x2 – y2| = |x|

(x2 – y2) = x

### Question 36(ii). xex/y – y + x(dy/dx) = 0, y(e) = 0

Solution:

We have,

xex/y – y + x(dy/dx) = 0

(dy/dx) = (y – xex/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) = (vx – xev)/x

v + x(dv/dx) = v – ev

x(dv/dx) = v – ev – v

x(dv/dx) = -ev

e-vdv = -(dx/x)

On integrating both sides,

âˆ«e-vdv = -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 – y2)dx – x2dy = 0, y(1) = 1

Solution:

We have,

(xy – y2)dx – x2dy = 0

(dy/dx) = (xy – y2)/x2

(dy/dx) = (y/x) – (y2/x2)

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 – v2

x(dv/dx) = v – v2 – v

x(dv/dx) = -v2

-(dv/v2) = (dx/x)

On integrating both sides,

-âˆ«(dv/v2) = 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 + 2v2 – 2v – v2)/(2 + v)

x(dv/dx) = (v2 – 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/v2| = log|xc|

(v – 1)3 = v2|xc|

(y – x)3/x3 = (y/x)2|xc|

(y – x)3 = y2x2c

At x = 1, y = 2,

(2 – 1)3 = 4 * 1 * c

c = (1/4)

(y – x)3 = (1/4)y2x2

### Question 36(vi). (y4 – 2x3y)dx + (x4 – 2xy3)dy = 0, y(1) = 0

Solution:

We have,

(y4 – 2x3y)dx + (x4 – 2xy3)dy = 0

dy/dx = (2x3y – y4)/(x4 – 2xy3)

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) = (2x3vx – v4x4)/(x4 – 2xv3x3)

v + x(dv/dx) = (2v – v4)/(1 – 2v3)

x(dv/dx) = [(2v – v4)/(1 – 2v3)] – v

x(dv/dx) = (2v – v4 – v + 2v4)/(1 – 2v3)

x(dv/dx) = (v + v4)/(1 – 2v3)

On integrating both sides,

âˆ«(dv/v) – âˆ«(3v2)dv/(1 + v3) = log|x| + log|c|

log|v| – log|1 + v3| = log|xc|

log|v/(1 + v3)| = log|xc|

At x = 1, y = 1,

1/(1 + 1) = c

c = (1/2)

(yx2)/(x3 + y3) = (1/2)x

### Question 36(vii). x(x2 + 3y2)dx + y(y2 + 3x2)dy = 0, y(1) = 1

Solution:

We have,

x(x2 + 3y2)dx + y(y2 + 3x2)dy = 0

dy/dx = -[x(x2 + 3y2)/y(y2 + 3x2)]

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 + 3v2 + v4 + 3v2)/v(v2 + 3)

[(v3 + 3v)/(1 + 6v2 + v4)]dv = -(dx/x)

Multiply both sides with 4 and integrating,

log|v4 + 6v2 + 1| = -log|x|4 + log|c|

|v4 + 6v2 + 1| = |c/x4|

(y4 + 6x2y2 + x4) = c

At y = 1, x = 1

(1 + 6 + 1) = c

c = 8

(y4 + 6x2y2 + x4) = 8

### Question 36(viii). {xsin2(y/x) – y}dx + xdy = 0, y(1) = Ï€/4

Solution:

We have,

{xsin2(y/x) – y}dx + xdy = 0

dy/dx = [y – xsin2(y/x)]/x

dy/dx = (y/x) – sin2(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 – sin2(v)

x(dv/dx) = v – sin2(v) – v

x(dv/dx) = -sin2(v)

-cosec2(v)dv = (dx/x)

On integrating both sides,

-âˆ«cosec2(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 + v2)/(1 – v)

(1 – v)dv/(1 + v + v2) = (dx/x)

On integrating both sides,

At x = 1, y = 0

âˆš3tan-1|1/âˆš3| – (1/2)log|1| = c

c = âˆš3(Ï€/6)

c = (Ï€/2âˆš3)

### Question 39. (dy/dx) = xy/(x2 + y2)

Solution:

We have,

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

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/(x2 + v2x2)

v + x(dv/dx) = v/(1 + v2)

x(dv/dx) = [v/(1 + v2)] – v

x(dv/dx) = (v – v – v3)/(1 + v2)

[-(1/v3) – (1/v)]dv = (dx/x)

On integrating both sides,

-âˆ«dv/v3 – âˆ«dv/v = âˆ«(dx/x)

(1/2v2) – log|v| = log|x| + c

(x2/2y2) = log|vx| + c

(x2/2y2) = log|(y/x)x| + c

(x2/2y2) = log|y| + c

At x = 0, y = 1

c = 0

(x2/2y2) = log|y|

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