Find the particular solution of the differential equation x2dy = (2xy + y2) dx, given that y = 1 when x = 1. - Mathematics

Find the particular solution of the differential equation x2dy = (2xy + y2) dx, given that y = 1 when x = 1.

Solution

x2dy = (2xy + y2)dx

=>dy/dx=(2xy+y^2)/x^2.......(i)

Let y=vx,

dy/dx=v+xdv/dx

Substituting in (i), we get

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

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

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

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

integrating both sides

=>int(dv)/(v^2+v)=intdx/x

=>(v+1-v)/(v(v+1))dv=intdx/x

=>logv-log|v+1|=logx+logC

=>log|v/(v+1)|=log|Cx|

=>log|(y/x)/(y/x+1)|=log|Cx|

=>y/(y+x)=Cx [Removing logarithm in both sides]

therefore y=Cxy+Cx^2 ,which is the general solution.

Putting y=1 and x=1,

1=C + C

=>2C=1

=>c=1/2y

=(xy)/2+x^2/2

therefore 2y=xy+x^2, which is the particular solution.

Concept: Methods of Integration: Integration by Substitution
Is there an error in this question or solution?