Answer to Question #128191 in Differential Equations for Sunil kumar meher

Question

(y-zx)p+(x+yz)q=x²+y²

Answer 1

$\frac{dx}{y-xz}=\frac{dy}{x+yz}=\frac{dz}{x^2+y^2}=t\\ dx=yt-xzt,\quad dy=xt+yzt,\quad dz=(x^2+y^2)t\\ ydx+xdy=y^2t-xyzt+x^2t+xyzt=(x^2+y^2)t\Longrightarrow\\ \frac{ydx+xdy}{x^2+y^2}=\frac{dz}{x^2+y^2}=t\Longrightarrow\\ dz=ydx+xdy\Longrightarrow z'_x=y,\quad z'_y=x \Longrightarrow z(x,y)=xy+c;\\ check:\\ (y-xz)p+(x+yz)q=(y-x\cdot (xy+c))\cdot z'_x + (x+y\cdot (xy+c))\cdot z'_y=\\ y^2-xy(xy+c)+x^2+xy(xy+c)=x^2+y^2;$