Transform the differential equation ${\frac{\partial^2z }{\partial x^2}}{x^2}-{\frac{\partial^2z }{\partial x \partial y}}{2xy}+{\frac{\partial^2z }{ \partial y^2}}{y^2}=0$ with the variable substitution $\left\{\begin{matrix} u=xy\\ v=\frac{1}{y} \end{matrix}\right.$.
Can you tell me where i should start with the problem?
EDIT: $$\frac{\mathrm{d}z }{\mathrm{d} x}=\frac{\partial z }{\partial u}y\\ \frac{\mathrm{d}z }{\mathrm{d} y}=\frac{\partial z }{\partial u}x+\frac{\partial z }{\partial v}\cdot (-\frac{1}{y^2})\\ $$
I'm having trouble to solve the second derivates.
EDIT: What I've tried without success. Deleted.
EDIT: $$\frac{\partial z}{\partial x}=y\frac{\partial z}{\partial u}\\ \frac{\partial z}{\partial y}=x\frac{\partial z}{\partial u}-\frac{1}{y^2}\frac{\partial z}{\partial v}\\ \frac{\partial^2 z}{\partial x^2}=y^2\frac{\partial^2 z}{\partial u^2}\\ \frac{\partial^2 z}{\partial x \partial y}=y(\frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial y } +\frac{\partial^2 z}{\partial u \partial v}\frac{\partial v}{\partial y })=y(x\frac{\partial^2z}{\partial u^2}-\frac{1}{y^2}\frac{\partial^2z}{\partial u \partial v})=xy\frac{\partial^2z}{\partial u^2}-\frac{1}{y}\frac{\partial^2z}{\partial u \partial v}\\ \frac{\partial^2 z}{\partial y^2}=x(\frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial y}+\frac{\partial^2 z}{\partial u \partial v}\frac{\partial v}{\partial y})-\frac{1}{y^2}(\frac{\partial^2 z}{\partial u \partial v}\frac{\partial u}{\partial y}+\frac{\partial^2 z}{\partial v^2}\frac{\partial v}{\partial y})\\ =x(x\frac{\partial^2 z}{\partial u^2}-\frac{1}{y^2}\frac{\partial^2 z}{\partial u \partial v})-\frac{1}{y^2}(x\frac{\partial^2 z}{\partial u \partial v}-\frac{1}{y^2}\frac{\partial^2 z}{\partial v^2})\\ =x^2\frac{\partial^2z}{\partial u^2}-\frac{2x}{y^2}\frac{\partial^2z}{\partial u \partial v}+\frac{1}{y^4}\frac{\partial^2z}{\partial v^2}$$
$$x^2(y^2\frac{\partial^2 z}{\partial u^2})-2xy(xy\frac{\partial^2z}{\partial u^2}-\frac{1}{y}\frac{\partial^2z}{\partial u \partial v})+y^2(x^2\frac{\partial^2z}{\partial u^2}-\frac{2x}{y^2}\frac{\partial^2z}{\partial u \partial v}+\frac{1}{y^4}\frac{\partial^2z}{\partial v^2})=0\\ \Rightarrow x^2y^2\frac{\partial^2 z}{\partial u^2}-2x^2y^2\frac{\partial^2 z}{\partial u^2}+2x\frac{\partial^2 z}{\partial u \partial v}+x^2y^2\frac{\partial^2 z}{\partial u^2}+\frac{1}{y^2}\frac{\partial^2z}{\partial v^2}-2x\frac{\partial^2 z}{\partial u \partial v}=0\\ \Rightarrow \frac{1}{y^2}\frac{\partial^2z}{\partial v^2}=0$$
I can't seem to remove $\frac{1}{y^2}\frac{\partial^2z}{\partial v^2}$ because it's the only one with $\frac{\partial^2z}{\partial v^2}$.
Is $\frac{1}{y^2}\frac{\partial^2z}{\partial v^2}=0$ the answer?
In fact $$ \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial z}{\partial u}y $$ and so on and on.
Update: \begin{eqnarray} \frac{\partial z}{\partial y}&=&\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}=x\frac{\partial z}{\partial u}-\frac{1}{y^2}\frac{\partial z}{\partial v},\\ \frac{\partial^2 z}{\partial x^2}&=&\frac{\partial}{\partial x}\frac{\partial z}{\partial x}=\frac{\partial}{\partial x}\frac{\partial z}{\partial u}y=y\bigg(\frac{\partial^2 z}{\partial u^2}\frac{\partial u}{\partial x}+\frac{\partial^2 z}{\partial u\partial v}\frac{\partial v}{\partial x}\bigg)=y^2\frac{\partial^2 z}{\partial u^2} \end{eqnarray} and so on and on.