Boundary Value Problem$ 5U_x+2U_y=(x-y)^2$ Solve the following boundary value problem
$5U_x+2U_y=(x-y)^2$
$U(3y,y)=\exp(y^3)$
Hint: Use change of variable $(x,y)\rightarrow(e,n)$ 
\begin{align}e &= 2x - 5y\\  n &= x - y\end{align}
I am struggling to understand why those boundaries have been picked, they do not seem to simplify the expression in any way. Usually one of the terms will cancel. Am I missing something?
 A: The hint is correct. Such change variables is really helpful:
$$
\begin{cases}
U_x=2U_e+U_n\\
U_y=-5U_e-U_n
\end{cases}
\quad\Rightarrow\quad
3U_n=n^2
\quad\Rightarrow\quad
U=\frac{n^3}{9}+f(e)
\quad\Rightarrow\quad
U(x,y)=f(2x-5y)+\frac{(x-y)^3}{9}
$$ 
Using the condition $U|_{x=3y}=e^{y^3}$ you can find function 
$f(y)=e^{y^3}-\frac{8y^3}{9}$, whence follows
$$
U(x,y)=e^{(2x-5y)^3}-\frac{8(2x-5y)^3}{9}+\frac{(x-y)^3}{9}\,.
$$
Remark.
To make the change of variables
$$
\begin{cases}
e=2x-5y\\
n=x-y
\end{cases}
$$
notice that 
$$
\begin{align}
U(x,y)=U\bigl(x(e,n),y(e,n)\bigr)=\widetilde{U}(e,n)\\
\widetilde{U}(e,n)=\widetilde{U}(e(x,y),n(x,y))=U(x,y)
\end{align}
$$
Hence follows
$$ 
\begin{cases}
U_x=\widetilde{U}_e e_x+\widetilde{U}_n n_x=2\widetilde{U}_e+\widetilde{U}_n\\
U_y=\widetilde{U}_e e_y+\widetilde{U}_n n_y=-5\widetilde{U}_e-\widetilde{U}_n
\end{cases}
$$
Formally, the tilde mark is needed. But two-and-a-half-century MathPhysics tradition
insists on omitting any tildes or their substitutes.
A: Hint: This is a general approach called the method of characteristics.
Inverting the variable transformation you have
$$x = (5n-e)/3, \\\ y = (2n-e)/3$$
Applying the change of variables let
$$\hat{U}(n,e)=U[x(n,e),y(n,e)]$$
Then 
$$\hat{U}_n= U_x\frac{\partial x}{\partial n}+U_y\frac{\partial y}{\partial n}=\frac{1}{3}(5U_x+2Uy) = \frac{1}{3}n^2$$
This can be integrated easily to find the general solution -- then apply the initial condition
A: You shall not pay attention to the hint but applying the so called method of characteristics for the present hyperbolic linear equation, which can be written in the form: 
$$f(x,y) \, u_x + g(x,y)\, u_y = h(x,y),$$ where $f$ and $g$ are constants and $h = (x-y)^2$. Then, $u$ is a constant over some curves called the characteristics, which are to be obtained from the equations:
$$ \frac{dx}{f} = \frac{dy}{g} = \frac{du}{h}. $$ The first equality tells us that, after substitution of your data, $2 dx - 5 dy  = 0$, therefore:
$$2x - 5y = \eta,$$ for some constant, $\eta$. To solve for $u$, you may take one of the two remaining equalities. Either $dx/f = du/h$ or $dy/g = du/h$. For example:
$$ \frac{du}{(x-y)^2} = 5^2 \frac{du}{(3x-\eta)^2}  =  \frac{dx}{5},   $$
which yields a separable equation for $u = u(x;\eta)$, whose solution is given by:
$$u(x;\eta)=  \alpha (3x-\eta)^3+ \beta, $$ with $\alpha$ a known value, while $\beta$ can be put as a function of $\eta$, so we would have the solution, substituting the value of $\eta$:
$$u(x,y) = \bar{\alpha} (x+5y)^3 + \beta(2x - 5y),$$  where $\bar{\alpha}$ is another known constant. If my math is correct, you just have to substitute your condition $u(3y,y) = e^{y^3}$ in order to obtain the unknown function $\beta$.
Cheers!
