Question

Dy/dx=2xy/x²+y² solve​

Answer 1

Answer:

$y=\frac{-1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}\ or \ \frac{1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}$

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

$\frac{dv}{dx}=\frac{-(v^2-1)v}{x(v^2+1)}$

$\frac{v^2+1}{(v^2-1)v}dv=\frac{-1}{x}dx$

∫$\frac{v^2+1}{(v^2-1)v}dv$ = ∫$\frac{-1}{x}dx$

u=v^2

du=2vdv

Left hand side:

∫$\frac{v^2+1}{v(v^2-1)}dv$

=∫$\frac{u+1}{2u(u-1)}du$

=$\frac{1}{2}$∫$(\frac{2}{u-1} -\frac{1}{u} )du$

=$ln(u-1)-\frac{ln(u)}{2} +c$

=$ln(v^2-1)-\frac{ln(v^2)}{2}+c$

Right hand side:

$=-ln(x)$

Solve for v:

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

$y=\frac{-1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}\ or \ \frac{1}{2} \sqrt[]{4x^2+4c_{1}^2} -c_{1}$