Answer to Question #128375 in Differential Equations for Sehba Ashraf

Question

Solve the PDE
(x+2z)p+(4zx-y)q=2x² +y

Answer 1

$\frac{dx}{x+2z}=\frac{dy}{4xz-y}=\frac{dz}{2x^2+y}=t\\ dx=xt+2zt,\quad dy=-yt+4xzt,\quad dz=2x^2t+yt\\ 2x\cdot dx-dy=2x^2t+4xzt+yt-4xzt=(2x^2+y)t\Rightarrow\\ dz=2x\cdot dx-dy \Rightarrow z=x^2-y+c;\\ Check:\\ (x+2z)p+(4xz-y)q=(x+2x^2-2y+2c)\cdot 2x+(4x^3-4xy+4cx-y)\cdot(-1)=\\ =2x^2+y;$