How can we apply the method of separation of variables? I want to check if the method of separation of variables can be used for the replacement of the following given partial differential equation from a pair of ordinary differential equations. If so, I want to find the equations.


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*$u_{xx}+(x+y) u_{yy}=0$


Suppose that $u$ is of the form $u(x,y)=X(x) Y(y)$.
Then  $u_{xx}+(x+y) u_{yy}=0 \Rightarrow X''(x) Y(y)+(x+y) X(x) Y''(y)=0 $.
So we see that we cannot use the method.
But in order to apply the method, we could set $z=x+y$.
But then how do we proceed? Do we find the derivative of z as or x?
EDIT:Let $z=x+y$. 
We have that $$\frac{dX}{dx}=\frac{dX}{dz}\cdot \frac{dz}{dx}=\frac{dX}{dz}$$ and $$\frac{d^2X}{dx^2}=\frac{d}{dx}\left (\frac{dX}{dz}\right )=\frac{d}{dz}\frac{dX}{dz}\cdot \frac{dz}{dx}=\frac{d^2X}{dz^2}$$ 
Then we have $$\frac{d^2X}{dx^2}\cdot Y+(x+y)\cdot X\cdot \frac{d^2Y}{dy^2}=0 \\ \Rightarrow \frac{d^2X}{dz^2}\cdot Y+z\cdot X\cdot \frac{d^2Y}{dy^2}=0 \\ \Rightarrow \frac{d^2X}{dz^2}\cdot Y=-z\cdot X\cdot \frac{d^2Y}{dy^2} \\ \Rightarrow \frac{1}{z\cdot X}\cdot \frac{d^2X}{dz^2}=- \frac{1}{Y}\cdot \frac{d^2Y}{dy^2}$$
But won't $X$ be a variable of both $y$ and $z$ since $x=z-y$?
Or do we get somehow that $X$ will depend only on $z$?
 A: $$u_{xx}+(x+y)u_{yy}=0$$
If you want change of variable such as $z=x+y$ without knowing more advanced method, you can do it with this very elementary process :
$$\text{Let}\quad\begin{cases}u(x,y)=U(x,z)\\z=x+y\quad\to\quad dz=dx+dy\end{cases}$$
$$du=u_x dx+u_y dy=U_x dx+U_z (dx+dy)=(U_x+U_z)dx+U_zdy\quad\to\quad\begin{cases}u_x=U_x+U_z\\u_y=U_z\end{cases}$$
$$du_x=u_{xx}dx+u_{xy}dy=dU_x+dU_z \quad \text{with}\quad \begin{cases} dU_x=U_{xx}dx+U_{xz}(dx+dy) \\ dU_z=U_{xz}dx+U_{zz}(dx+dy)\end{cases}$$
$$u_{xx} dx+u_{xy} dy=(U_{xx}+2U_{xz}+U_{zz})dx+(U_{xz}+U_{zz})dy\quad\to\quad$$
$$ u_{xx}=U_{xx}+2U_{xz}+U_{zz}$$
$$du_y=u_{xy}dx+u_{yy}dy=dU_z=U_{xz}dx+U_{zz}(dx+dy)=(U_{xz}+U_{zz})dx+U_{zz}dy \quad\to\quad u_{yy}=U_{zz}$$
$$u_{xx}+(x+y)u_{yy}=0\quad\to\quad U_{xx}+2U_{xz}+U_{zz}+z U_{zz}=0 $$
$$U_{xx}+2U_{xz}+(z+1)U_{zz}=0$$
Then, the transformed PDE can be separated with $\quad U(x,z)=X(x)Z(z)$
$$\frac{X''}{X}+2\frac{X'}{X}\frac{Z'}{Z}+(z+1)\frac{Z''}{Z}=0$$
With $\quad \frac{X'}{X}=\lambda=\text{constant}\quad\to\quad X=e^{\lambda x}\quad\to\quad \lambda^2+2\lambda \frac{Z'}{Z}+(z+1)\frac{Z''}{Z}=0$
$$(z+1)Z''+2\lambda Z'+\lambda^2 Z=0$$
The solution involves Bessel functions.
Note : Don't forget that the separation of variables method doesn't give the general solution of the PDE, but only particular solutions (one for each value of $\lambda$ , with an arbitrary coefficient for each term). In order to obtain more general solutions, one have to linearly add those particular solutions : Either on form of series with discret values of $\lambda$ , or on continuous form involving an integral with respect to $\lambda$ and where the arbitrary coefficients are replaced by an arbitrary function of $\lambda$.
A: Hint:
Let $\begin{cases}w=x\\z=x+y\end{cases}$ ,
Then $u_x=u_ww_x+u_zz_x=u_w+u_z$
$u_{xx}=(u_w+u_z)_x=(u_w+u_z)_ww_x+(u_w+u_z)_zz_x=u_{ww}+u_{wz}+u_{wz}+u_{zz}=u_{ww}+2u_{wz}+u_{zz}$
$u_y=u_ww_y+u_zz_y=u_z$
$u_{yy}=(u_z)_y=(u_z)_ww_y+(u_z)_zz_y=u_{zz}$
$\therefore u_{ww}+2u_{wz}+(z+1)u_{zz}=0$
$u_{xx}+2u_{xz}+(z+1)u_{zz}=0$
