Question

Find the particular solution of the differential equation `dy/dx=(xy)/(x^2+y^2)` given that y = 1, when x = 0.

Answer 1

`dy/dx=(xy)/(x^2+y^2)                 .....(1)`

This is a homogenous differential equation.

Substitute y = vx             .....(2)

`⇒dy/dx=v+x (dv)/dx            .....(3)`

From (1), (2) and (3), we have

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

`⇒x (dv)/dx+v=v/(1+v^2)`

`⇒x (dv)/dx=v/(1+v^2)-v=(v-v-v^2)/(1+v^2)`

`⇒x (dv)/dx=−v^3/(1+v^2)`

`⇒(1+v^2)/v^3dv=−dx/x`

`⇒(1/v^3+1/v)dv=−dx/x`

Integrating both sides, we have

`v^(−3+1)/(−3+1)+lnv=−lnx+C`

`⇒−1/(2v^2)+lnv=−lnx+C`

`⇒−1/(2v^2)+lnvx=C`

`⇒−x^2/(2y^2)+lny=C`

Given: y = 1 when x = 0

⇒ C = 0

Thus, the particular solution of the given differential equation is given by

`lny=x^2/(2y^2)`

or x² = 2y² lny