Class 12 RD Sharma Solutions β Chapter 22 Differential Equations β Exercise 22.9 | Set 3
Question 27. (x2 β 2xy)dy + (x2 β 3xy + 2y2)dx = 0
Solution:
We have,
(x2 β 2xy)dy + (x2 β 3xy + 2y2)dx = 0
(dy/dx) = (x2 β 3xy + 2y2)/(2xy β 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 β 3xvx + 2v2x2)/(2xvx β x2)
v + x(dv/dx) = (1 β 3v + 2v2)/(2v β 1)
x(dv/dx) = [(1 β 3v + 2v2)/(2v β 1)] β v
x(dv/dx) = (1 β 3v + 2v2 β 2v2 + v)/(2v β 1)
x(dv/dx) = (1 β 2v)/(2v β 1)
x(dv/dx) = -1
dv = -(dx/x)
On integrating both sides,
β«dv = -β«(dx/x)
v = -log|x| + log|c|
(y/x) + log|x| = log|c| (Where βcβ is integration constant)
Question 28. x(dy/dx) = y β xcos2(y/x)
Solution:
We have,
x(dy/dx) = y β xcos2(y/x)
(dy/dx) = y/x β 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 β cos2(v)
x(dv/dx) = -cos2(v)
dv/cos2(v) = -(dx/x)
On integrating both sides,
β«dv/cos2(v) = -β«(dx/x)
β«sec2vdv = -β«(dx/x)
tan(v) = -log|x| + log|c|
tan(y/x) = log|c/x| (Where βcβ is integration constant)
Question 29. x(dy/dx) β y = 2β(y2 β x2)
Solution:
We have,
x(dy/dx) β y = 2β(y2 β x2)
(dy/dx) = [2β(y2 β 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) = 2β(v2 β 1)
dv/β(v2 β 1) = 2(dx/x)
On integrating both sides,
β«dv/β(v2 β 1) = 2β«(dx/x)
log|v + β(v2β 1)| = 2log(x) + log(c)
|v + β(v2 β 1)| = |cx2|
(Where βcβ is integration constant)
Question 30. xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy β ydx)
Solution:
We have,
xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy β ydx)
xycos(y/x)dx + x2cos(y/x)dy = xysin(y/x)dy β y2sin(y/x)dx
x2cos(y/x)dy β xysin(y/x)dy = -y2sin(y/x)dx β 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 β cosv)
x(dv/dx) = [(vcosv + v2sinv)/(vsinv β cosv)] β v
x(dv/dx) = (vcosv + v2sinv β v2sinv + vcosv)/(vsinv β cosv)
x(dv/dx) = (2vcosv)/(vsinv β cosv)
[(vsinv β cosv)/(vcosv)]dv = 2(dx/x)
On integrating both sides,
β«tanvdv β β«(dv/v) = 2log|x| + log|c|
log|secv| β log|v| = log|cx2|
log|(secv/v)| = log|cx2|
(x/y)sec(y/x) = cx2
sec(y/x) = cxy (Where βcβ is integration constant)
Question 31. (x2 + 3xy + y2)dx β x2dy = 0
Solution:
We have,
(x2 + 3xy + y2)dx β 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) β v
x(dv/dx) = (1 + 2v + v2)
x(dv/dx) = (1 + v)2
dv/(1 + v)2 = (dx/x)
On integrating both sides,
β«dv/(1 + v)2 = β«(dx/x)
-[1/(v + 1)] = log|x| β c
x/(x + y) + log|x| = c (Where βcβ is an integration constant)
Question 32. (x β y)(dy/dx) = (x + 2y)
Solution:
We have,
(x β y)(dy/dx) = (x + 2y)
(dy/dx) = (x + 2y)/(x β 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 β vx)
v + x(dv/dx) = (1 + 2v)/(1 β v)
x(dv/dx) = [(1 + 2v)/(1 β v)] β v
x(dv/dx) = (1 + 2v β v + v2)/(1 β v)
x(dv/dx) = (1 + v + v2)/(1 β v)
(1 β v)dv/(1 + v + v2) = (dx/x)
On integrating both sides,
β«[(1 β v)/(1 + v + v2)]dv = β«(dx/x)
(Where βcβ is an integration constant)
Question 33. (2x2y + y3)dx + (xy2 β 3x2)dy = 0
Solution:
We have,
(2x2y + y3)dx + (xy2 β 3x2)dy = 0
dy/dx = (2x2y + y3)/(3x3 β 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 β xv2x2)
v + x(dv/dx) = (2v + v3)/(3 β v3)
x(dv/dx) = [(2v + v3)/(3 β v3)] β v
x(dv/dx) = (2v + v3 β 3v + v3)/(3 β v3)
(3 β v3)dv/(2v3 β v) = (dx/x)
On integrating both sides,
β«[(3 β v3)/(2v3 β v)]dv = β«(dx/x)
Using partial fraction,
3 β v2 = A(2v2 β 1) + (Bv + C)v
3 β v2 = 2Av2 β A + Bv2 + Cv
3 β v2 = v2(2A + B) + Cv β 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 β x2|5 = x2c4y12 (Where βcβ is an integration constant)
Question 34. x(dy/dx) β y + xsin(y/x) = 0
Solution:
We have,
x(dy/dx) β y + xsin(y/x) = 0
x(dy/dx) = y β xsin(y/x)(dy/dx) = [y β 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 β xsinv]/x
v + x(dv/dx) = (v β sinv)
x(dv/dx) = -sinv
cosecvdv = -(dx/x)
On integrating both sides,
β«cosecvdv = -β«(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 βcβ is integration constant)
Question 35. ydx + {xlog(y/x)}dy β 2xydy = 0
Solution:
We have,
ydx + {xlog(y/x)}dy β 2xydy = 0
y + {xlog(y/x)}(dy/dx) β 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 β logv)
x(dv/dx) = [v/(2 β logv)] β v
x(dv/dx) = (v β 2v + vlogv)/(2 β logv)
x(dv/dx) = -v(logv β 1)/(logv β 2)
On integrating both sides,
Let. logv β 1 = z
On differentiating both sides,
(dv/v) = dz
β«dz β β«(dz/z) = -β«(dx/x)
z β log|z| = -log|x| + log|c|
(logv β 1) β log|(logv β 1)| = -log|x| + log|c|
logv β log|logv β 1| = -log|x| + log|c| + 1
log|(logv β 1)/v| = log|c1x|
|logv β 1| = |c1xv|
|log(y/x) β 1| = |c1x(y/x)|
|log(y/x) β 1| = |c1y| (Where βc1β 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] β v
x(dv/dx) = (1 + v2 β 2v2)/2v
x(dv/dx) = (1 β v2)/2v
2vdv/(1 β v2) = (dx/x)
On integrating both sides,
β«2vdv/(1 β v2) = β«(dx/x)
-log|1 β v2| = log|x| β log|c|
log|1 β v2| = log|c/x|
|1 β y2/x2| = |c/x|
|x2 β y2| = |cx|
At x = 1, y = 0
1 β 0 = c
c = 1
|x2 β y2| = |x|
(x2 β y2) = x
Question 36(ii). xex/y β y + x(dy/dx) = 0, y(e) = 0
Solution:
We have,
xex/y β y + x(dy/dx) = 0
(dy/dx) = (y β 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 β xev)/x
v + x(dv/dx) = v β ev
x(dv/dx) = v β ev β v
x(dv/dx) = -ev
e-vdv = -(dx/x)
On integrating both sides,
β«e-vdv = -β«(dx/x)
-e-v = -log|x| β 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) β (y/x) + cosec(y/x) = 0, y(1) = 0
Solution:
We have,
(dy/dx) β (y/x) + cosec(y/x) = 0
(dy/dx) = (y/x) β 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 β cosec(v)
x(dv/dx) = v β cosec(v) β v
x(dv/dx) = -cosec(v)
-sin(v)dv = (dx/x)
On integrating both sides,
-β«sin(v)dv = β«(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) β 1
Question 36(iv). (xy β y2)dx β x2dy = 0, y(1) = 1
Solution:
We have,
(xy β y2)dx β x2dy = 0
(dy/dx) = (xy β y2)/x2
(dy/dx) = (y/x) β (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 β v2
x(dv/dx) = v β v2 β v
x(dv/dx) = -v2
-(dv/v2) = (dx/x)
On integrating both sides,
-β«(dv/v2) = β«(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] β v
x(dv/dx) = (v + 2v2 β 2v β v2)/(2 + v)
x(dv/dx) = (v2 β v)/(2 + v)
(2 + v)dv/[v(v β 1)] = (dx/x)
On integrating both sides,
Using partial derivative,
2 + v = A(v β 1) + B(v)
2 + v = v(A + B) β A
On comparing the coefficients,
A = -2
B = 3
-2β«(dv/v) + 3β«dv/(v β 1) = β«(dx/x)
-2log|v| + 3log|v β 1| = log|x| + log|c|
log|(v β 1)3/v2| = log|xc|
(v β 1)3 = v2|xc|
(y β x)3/x3 = (y/x)2|xc|
(y β x)3 = y2x2c
At x = 1, y = 2,
(2 β 1)3 = 4 * 1 * c
c = (1/4)
(y β x)3 = (1/4)y2x2
Question 36(vi). (y4 β 2x3y)dx + (x4 β 2xy3)dy = 0, y(1) = 0
Solution:
We have,
(y4 β 2x3y)dx + (x4 β 2xy3)dy = 0
dy/dx = (2x3y β y4)/(x4 β 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 β v4x4)/(x4 β 2xv3x3)
v + x(dv/dx) = (2v β v4)/(1 β 2v3)
x(dv/dx) = [(2v β v4)/(1 β 2v3)] β v
x(dv/dx) = (2v β v4 β v + 2v4)/(1 β 2v3)
x(dv/dx) = (v + v4)/(1 β 2v3)
On integrating both sides,
β«(dv/v) β β«(3v2)dv/(1 + v3) = log|x| + log|c|
log|v| β 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) β y}dx + xdy = 0, y(1) = Ο/4
Solution:
We have,
{xsin2(y/x) β y}dx + xdy = 0
dy/dx = [y β xsin2(y/x)]/x
dy/dx = (y/x) β 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 β sin2(v)
x(dv/dx) = v β sin2(v) β v
x(dv/dx) = -sin2(v)
-cosec2(v)dv = (dx/x)
On integrating both sides,
-β«cosec2(v) = β«(dx/x)
cot(v) = log|x| + log|c|
cot(y/x) = log|xc|
At x = 1, y = Ο/4
cot(Ο/4) = log|c|
log|c| = 1
c = e
cot(y/x) = log|ex|
Question 36(ix). x(dy/dx) β y + xsin(y/x) = 0, y(2) = Ο
Solution:
We have,
x(dy/dx) β y + xsin(y/x) = 0
x(dy/dx) = y β xsin(y/x)
(dy/dx) = (y/x) β 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 β sin(v)
x(dv/dx) = v β sin(v) β v
x(dv/dx) = -sin(v)
-cosec(v)dv = (dx/x)
On integrating both sides,
β«cosec(v) = -β«(dx/x)
log|cosec(v) β cot(v)| = -log|x| + log|c|
log|cosec(v) β cot(v)| = -log|x| + log|c|
log|cosec(y/x) β cot(y/x)| = -log|x| + log|c|
At x = 2, y = Ο
|cosec(Ο/2) β cot(Ο/2)| = -log|2| + log|c|
log|c| = 0
log|cosec(y/x) β cot(y/x)| = -log|x|
Question 37. xcos(y/x)(dy/dx) = ycos(y/x) + x, When x = 1, y = Ο/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 β v
x(dv/dx) = 1/cosv
cosvdv = (dx/x)
On integrating both sides,
β«cosvdv = β«(dx/x)
sin(v) = log|x| + log|c|
sin(y/x) = log|x| + log|c|
At x = 1, y = Ο/4
1/β2 = 0 + log|c|
log|c| = (1/β2)
sin(y/x) = log|x| + (1/β2)
Question 38. (x β y)(dy/dx) = (x + 2y), when x = 1,y = 0
Solution:
We have,
(x β y)(dy/dx) = (x + 2y)
(dy/dx) = (x + 2y)/(x β 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 β vx)
v + x(dv/dx) = (1 + 2v)/(1 β v)
x(dv/dx) = [(1 + 2v)/(1 β v)] β v
x(dv/dx) = (1 + 2v β v + v2)/(1 β v)
(1 β v)dv/(1 + v + v2) = (dx/x)
On integrating both sides,
At x = 1, y = 0
β3tan-1|1/β3| β (1/2)log|1| = c
c = β3(Ο/6)
c = (Ο/2β3)
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)] β v
x(dv/dx) = (v β v β v3)/(1 + v2)
[-(1/v3) β (1/v)]dv = (dx/x)
On integrating both sides,
-β«dv/v3 β β«dv/v = β«(dx/x)
(1/2v2) β 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|