$(x^2+3y^2+3u^2)u_x-2xyu_y+2xu=0$
This is my solution;
${\frac {dx}{x^2+3y^2+3u^2}}$=${\frac {dy}{-2xy}}$=${\frac {du}{-2xu}}$= $λ$
${\frac {dy}{-2xy}}$=${\frac {du}{-2xu}}$
${\frac {du}{u}}$-${\frac {dy}{y}}$=$0$
$lnu-lny=lnc_1$
${\frac {u}{y}}$=$c_1$
then i know that we now need to find another function. but i am stuck.
$F(c_1,c_2)=0$
$c_1=f(c_2)$
can u help me to find $c_2$ with steps, please? i would appreciate it if you help.
$2xy dx + (x^2+3y^2+\dfrac{3}{c^2}y^2) dy = {d\phi}= 0$
${d\phi}=c_2$
${\phi}(x,y,u)=x^2y+y^3+yu^2$=$c_2$