prove $z_{xx}-\frac{1}{x}z_x=4x^2z_{yy}$

Question

assume $z$ is a function of $x,y$ and $z=f(x^2-y)+g(x^2+y)$ prove $z_{xx}-\frac{1}{x}z_x=4x^2z_{yy}$

assume $x^2-y=u,x^2+y=v$ using chain rule we have : $z_x=2x(z_u+z_v)$

so $z_{xx}=2x(\frac{1}{x}(z_u+z_v)+2x(z_{uu}+z_{vv}))+2x(\frac{1}{x}(z_u+z_v)+2x(z_{uv}+z_{vv}))= 4(z_u+z_v)+4x^2(z_{uu}+z_{vv}+z_{uv}+z_{vu}) $ similarly we have :$z_y=-z_u+z_v$

so $z_{yy}=z_{uu}-z_{vu}-z_{vu}+z_{vv}$

there for replacing in the given equation we have to prove : $2(z_u+z_v)+8x^2(z_{uv}+z_{vu})=0$

but I'm stuck here ...

any help is appreciated , thanks!

Answer 1

Just work with $f$ and $g$ directly. We have

$$z_x=2x(f'(x^2-y)+g'(x^2+y)),$$ $$z_{xx}=2(f'(x^2-y)+g'(x^2+y))+4x^2(f''(x^2-y)+g''(x^2+y)),$$

and so $$z_{xx}-\frac{1}{x}z_x=4x^2(f''(x^2-y)+g''(x^2+y)),$$

which is exactly $4x^2z_{yy}.$