# Solution of the PDE?

#### bkarpuz

##### New member
Dear MHB members,

I have the following equation
$xy(z_{xx}-z_{yy})+(x^{2}-y^{2})z_{xy}=yz_{x}-xz_{y}-2(x^{2}-y^{2})$.
When I transform this into the canonical form via $\xi=2xy$ and $\eta=x^{2}-y^{2}$, I obtain
$z_{\xi\eta}+\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}=-\frac{\eta}{2(\xi^{2}+\eta^{2})}$.
This is the point I stuck at.
It follows from here that
$z_{\xi}=\frac{f_{1}(\xi)}{\sqrt{\xi^{2}+\eta^{2}}}-\frac{1}{2}$,
which yields
$z=\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u+f_{2}(\eta)-\frac{\xi}{2}$.
Therefore, the given equation has the general solution
$z=\int^{2xy}\frac{f_{1}(u)}{\sqrt{u^{2}+(x^{2}-y^{2})^{2}}}{\rm d}u+f_{2}(x^{2}-y^{2})-xy$.

My question is how to modify the part
$\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u$ in a better way
and get rid of the integral if possible.

Thank you very much.
bkarpuz

#### Sudharaka

##### Well-known member
MHB Math Helper
Dear MHB members,

I have the following equation
$xy(z_{xx}-z_{yy})+(x^{2}-y^{2})z_{xy}=yz_{x}-xz_{y}-2(x^{2}-y^{2})$.
When I transform this into the canonical form via $\xi=2xy$ and $\eta=x^{2}-y^{2}$, I obtain
$z_{\xi\eta}+\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}=-\frac{\eta}{2(\xi^{2}+\eta^{2})}$.
This is the point I stuck at.
It follows from here that
$z_{\xi}=\frac{f_{1}(\xi)}{\sqrt{\xi^{2}+\eta^{2}}}-\frac{1}{2}$,
which yields
$z=\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u+f_{2}(\eta)-\frac{\xi}{2}$.
Therefore, the given equation has the general solution
$z=\int^{2xy}\frac{f_{1}(u)}{\sqrt{u^{2}+(x^{2}-y^{2})^{2}}}{\rm d}u+f_{2}(x^{2}-y^{2})-xy$.

My question is how to modify the part
$\int^{\xi}\frac{f_{1}(u)}{\sqrt{u^{2}+\eta^{2}}}{\rm d}u$ in a better way
and get rid of the integral if possible.

Thank you very much.
bkarpuz
Hi bkarpuz, I don't have any idea about how you obtained $$z_{\xi}=\frac{f_{1}(\xi)}{\sqrt{\xi^{2}+\eta^{2}}}-\frac{1}{2}$$. However if you integrate, $$z_{\xi\eta}+\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}=-\frac{\eta}{2(\xi^{2}+\eta^{2})}$$ with respect to $$\eta$$ we can obtain,

$z_{\xi}+\int\frac{\eta}{\xi^{2}+\eta^{2}}z_{\xi}\,d\eta=-\frac{1}{4}\ln(\xi^{2}+\eta^{2})+g(\xi)$

Kind Regards,
Sudharaka.