Answer:
[tex]y=\frac{-1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}\ or \ \frac{1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}[/tex]
Step-by-step 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
[tex]\frac{dv}{dx}=\frac{-(v^2-1)v}{x(v^2+1)}[/tex]
[tex]\frac{v^2+1}{(v^2-1)v}dv=\frac{-1}{x}dx[/tex]
u=v^2
du=2vdv
Left hand side:
∫[tex]\frac{v^2+1}{v(v^2-1)}dv[/tex]
=∫[tex]\frac{u+1}{2u(u-1)}du[/tex]
=[tex]\frac{1}{2}[/tex]∫[tex](\frac{2}{u-1} -\frac{1}{u} )du[/tex]
=[tex]ln(u-1)-\frac{ln(u)}{2} +c[/tex]
=[tex]ln(v^2-1)-\frac{ln(v^2)}{2}+c[/tex]
Right hand side:
[tex]=-ln(x)[/tex]
Solve for v:
[tex]v=-\frac{-c_{1}+\sqrt{c_{2}+4x^2 } }{2x} \ or \ \frac{-c_{1}+\sqrt{c_{2}+4x^2 } }{2x}\\[/tex]
[tex]y=\frac{-1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}\ or \ \frac{1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}[/tex]
