Why solving $\dfrac{\partial u}{\partial x}=\dfrac{\partial^2u}{\partial y^2}$ like this is wrong? Try let $v=x+y$ , $w=x-y$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial x}=\dfrac{\partial u}{\partial w}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial y}=-\dfrac{\partial u}{\partial w}$
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(-\dfrac{\partial u}{\partial w}\right)=-\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial y}=-\dfrac{\partial^2u}{\partial vw}$
$\therefore-\dfrac{\partial^2u}{\partial vw}=\dfrac{\partial u}{\partial w}$
Let $z=\dfrac{\partial u}{\partial w}$ ,
Then $\dfrac{\partial z}{\partial v}=\dfrac{\partial^2u}{\partial vw}$
$\therefore-\dfrac{\partial z}{\partial v}=z$
$\dfrac{dz}{z}=-~dv$
$\int\dfrac{dz}{z}=\int-~dv$
$\ln z=-v+c_1(w)$
$z=c_2(w)e^{-v}$
$\dfrac{\partial u}{\partial w}=c_2(w)e^{-v}$
$u=\int c_2(w)e^{-v}~dw$
$u=C_1(v)+C_2(w)e^{-v}$
$u=C_1(x+y)+C_2(x-y)e^{-x-y}$
I did it correctly in Finding an analytical solution to the wave equation using method of characteristics , and did it not known whether correct or not in Solving $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$ , but why in here is wrong?
 A: Fleshing out @QiaochuYuan's comment a little: 
$$\begin{eqnarray*}
\frac{\partial}{\partial x}
&=& \frac{\partial v}{\partial x} \frac{\partial}{\partial v}
+ \frac{\partial w}{\partial x} \frac{\partial}{\partial w} \\
&=& \frac{\partial}{\partial v} + \frac{\partial}{\partial w} \\
\frac{\partial}{\partial y}
&=& \frac{\partial v}{\partial y} \frac{\partial}{\partial v}
+ \frac{\partial w}{\partial y} \frac{\partial}{\partial w} \\
&=& \frac{\partial}{\partial v} - \frac{\partial}{\partial w}. \\
\end{eqnarray*}$$
The reason this is a nice change of variables for the wave equation is because
$$\frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} 
= 4 \frac{\partial}{\partial v} \frac{\partial}{\partial w}.$$
No such nice thing happens with the heat equation. 
Addendum: Note that 
$$\begin{eqnarray*}
\frac{\partial^2}{\partial x^2}
    &=&  \left(\frac{\partial}{\partial v} + \frac{\partial}{\partial w}\right)^2 \\
    &=& \frac{\partial^2}{\partial v^2} 
    + 2 \frac{\partial}{\partial v} \frac{\partial}{\partial w}
    + \frac{\partial^2}{\partial w^2} \\
\frac{\partial^2}{\partial y^2}
    &=&  \left(\frac{\partial}{\partial v} - \frac{\partial}{\partial w}\right)^2 \\
    &=& \frac{\partial^2}{\partial v^2}
    - 2 \frac{\partial}{\partial v} \frac{\partial}{\partial w}
    + \frac{\partial^2}{\partial w^2}. \\    
\end{eqnarray*}$$
The transformed equation is 
$$\left(\frac{\partial}{\partial v} + \frac{\partial}{\partial w}\right)u(v,w)
= \left(\frac{\partial^2}{\partial v^2}
    - 2 \frac{\partial}{\partial v} \frac{\partial}{\partial w}
    + \frac{\partial^2}{\partial w^2}\right) u(v,w).$$
There are no nice cancelations.
This transformation just makes the PDE harder to solve.
A: This equation is a heat equation.
Usually we use the methods of seperation of variables or Fourier transforms.
Seperation of variables:
Let $$ u=X(x)*Y(y),$$ then we get $$X'(x)Y(y)=X(x)Y''(y).$$
That is to say, $$X'/X=Y''/Y$$
Notice that the left side of the equation above is a function of $x$, while the right side the function of $y$. So if the left side equals the right side, then both of them must be constant.
Thus
$$X'/X=c,Y''/Y=c,$$
these are ODEs which are easy to solve.
Fourier Transform:
Notice that we use Fourier Transform with respect of $y.$
