Answer to Question #283606 in Differential Equations for Thor

Question

solve the following differential equation: 2x^2y(d^2y/dx^2)+4y^2=x^2(dy/dx)^2+2xy(dy/dx)

Answer 1

$2x^2yy''+4y^2=x^2(y')^2+2xyy'$

$y=x^2z^2$

then:

$y=x^2z^2$

$y'=2xz^2+2x^2zz'$

$y''=2z^2+4xzz'+4xzz'+2x^2((z')^2+zz'')$

$2x^2x^2z^2(2z^2+4xzz'+4xzz'+2x^2((z')^2+zz''))+4x^4z^4=$

$=x^2(2xz^2+2x^2zz')^2+2xx^2z^2(2xz^2+2x^2zz')$

$x^2z^2(z^2+4xzz'+x^2((z')^2+zz''))+x^2z^4=$

$=x^2z^4+2x^3z^3z'+x^4z^2(z')^2+xz^2(xz^2+x^2zz')$

$z^2+4xzz'+x^2((z')^2+zz'')+z^2=z^2+2xzz'+x^2(z')^2+z^2+xzz'$

$xzz'+x^2zz''=0$

$xz(z'+xz'')=0$

$z'+xz''=0$

$z'=u$

$u+xu'=0$

$du/u=-dx/x$

$lnu=-lnx+lnc_1$

$u=c_1/x$

$dz=c_1dx/x$

$z=c_1lnx+c_2$

$y(x)=x^2(c_1lnx+c_2)^2$