I have this boundary conditions question:
$2\frac{\partial u}{\partial x}-14\frac{\partial u}{\partial y}=3(x+y)$
Where the boundary conditions are:
$u(x,-6x)=Cos((2x)^2) \quad \forall x \in \mathbb{R}$
I don't understand the boundary conditions. Also, what would a proper substitution of variables be?
The Boundary value gives you value of $u(x,y)$ on the line $y=-6x$ or curve $(t,-6t)$. you can try to introduce two different variables to make the equation integrable. like $$ \left\{ \begin{array}{} \xi = a_1 x+a_2y \\ \eta = b_1x+b_2y \end{array} \right. \tag{1} \label{1} $$ you can easily guess $\xi$ as $x+y$, because of RHS (we want the PDE to be integrable ODE after substitution). and for second one insert $\eqref{1}$ in PDE $$ (2-14) \frac{\partial u}{\partial \xi} + \underbrace{(2b_1-14b_2)}_{\text{Want to be 0}} \frac{\partial u}{\partial \eta} = 3 \xi $$ so we can choose $$ b_1 = 7 , b_2 = 1 \rightarrow \eta=7x+y $$ therefore the PDE becomes: $$ \frac{\partial u}{\partial \xi} = -\frac{1}{4}\xi \Rightarrow u=-\frac{1}{8}\xi^2 + g(\eta) $$ Finally we get $$ u(x,y) = -\frac{1}{8} (x+y)^2 + g(7x+y). \tag{2}\label{2} $$ now you can use boundary condition $$ u(x,-6x) = -\frac{1}{8} (5x)^2 + g(x) = \cos(4x^2) $$ hence : $$ g(z) = \cos(4z^2)+\frac{25}{8} x^2 \\ \Rightarrow g(7x+y) = \cos(4(7x+y)^2)+\frac{25}{8} (7x+y)^2. $$ put this back into $\eqref{2}$ and you're done.