Answer to Question #275453 in Differential Equations for Jadeey

Question

2x^ 3 y^ prime =y(y^ 2 +3x^ 2 )

Answer 1

$2x^ 3 y' =y(y^ 2 +3x^ 2 )$

$y=tx$

$y'=t'x+t$

$2x^ 3 (t'x+t) =tx((tx)^ 2 +3x^ 2 )$

$2x^4t'+2x^3t=t^3x^3+3tx^3$

$2xt'+2t=t^3+3t$

$\frac{dt}{t^3+t}=\frac{dx}{2x}$

$-\frac{ln(1/t^2+1)}{2}=\frac{lnx}{2}+lnc/2$

$\frac{1}{1/t^2+1}=cx$

$\frac{y^2}{x^2+y^2}=cx$