Question

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

Answer 1

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.