Question

Dy/dx=2xy/x²+y² solve​

Answer 1

`y=(-1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)\ or \ (1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)`

Explanation:

As it is first order nonlinear ordinary differential equation

Let y(x) = x v(x)

2xy/(x²+y²)=2v/(v^2+1)

dy=xdv+vdx

dy/dx=d(dv/dx)+v

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

dv/dx=[(2v)/(v^2+1)-v]/x

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

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

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

u=v^2

du=2vdv

Left hand side:

∫
`(v^2+1)/(v(v^2-1))dv`

=∫
`(u+1)/(2u(u-1))du`

=
`(1)/(2)`∫
`((2)/(u-1) -(1)/(u) )du`

=
`ln(u-1)-(ln(u))/(2) +c`

=
`ln(v^2-1)-(ln(v^2))/(2)+c`

Right hand side:

`=-ln(x)`

Solve for v:

`v=-\frac{-c_(1)+\sqrt{c_(2)+4x^2 } }{2x} \ or \ \frac{-c_(1)+\sqrt{c_(2)+4x^2 } }{2x}\\`

`y=(-1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)\ or \ (1)/(2) \sqrt[]{4x^2+4c_(1)^2} -c_(1)`