# PDE1

$3u_x+4u_y-2u=1, u(x,0)=e^x\,$

Introduce the change of coordinates

$\xi = A x + B y \,$
$\eta = C x + D y \,$

So $u\,$ is now a function of $(\xi,\eta)\,$. Recompute the derivatives with respect to these new coordinates.

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x} + \frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial x} = A u_\xi\ + C u_\eta\,$,

$\frac{\partial u}{\partial y} = \frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial y} + \frac{\partial u}{\partial \eta} \frac{ \partial \eta}{\partial y} = B u_\xi + D u_\eta\,$.

Plug these derivatives into the original equation.

$3(Au_\xi + Cu_\eta) + 4(Bu_\xi + Du_\eta) - 2u = 1\,$

$u_\xi(3A+4B) + u_\eta(3C+4D)-2u=1\,$

Choose any values of A,B,C,D to make the problem easy to solve, with the caveat that $AD-BC \ne 0$. For example, let C = 1,D = − 3 / 4 so that the uη term goes away. Then let A = 1 / 3,B = 0 so that the uξ coefficient equals unity. Then,

$u_\xi-2u=1\,$

Use the integrating factor $e^{\int -2 d\xi} = e^{-2\xi}\,$

$e^{-2\xi}u_\xi - e^{-2\xi}2u=e^{-2\xi}\,$

$\frac{\partial}{\partial \xi}\left[e^{-2\xi}u\right]=e^{-2\xi}\,$

$e^{-2\xi}u = \frac{-1}{2}e^{-2\xi}+f(\eta)\,$

Here $f(\eta)\,$ is an undetermined function of η.

$u(\xi,\eta) = \frac{-1}{2}+e^{2\xi}f(\eta)\,$

$u(x,y) = \frac{-1}{2} + e^{\frac{2}{3}x}f(x-\frac{3}{4}y)\,$

Now we determine the unique solution that corresponds to the inital condition $u(x,0)=e^x\,$. Using the general solution above, set y = 0.

$u(x,0) = \frac{-1}{2} + e^{\frac{2}{3}x}f(x)=e^x\,$

$f(x) = \frac{e^x+\frac{1}{2}}{e^{\frac{2}{3}x}}\,$

But in the general solution, we have $f(x-\frac{3}{4}y)$ instead of $f(x)\,$. So making the appropriate substitutions,

$f(x-\frac{3}{4}y) = \frac{e^{x-\frac{3}{4}y}+\frac{1}{2}}{e^{\frac{2}{3}(x-\frac{3}{4}y)}}\,$

Plug this function into the general solution.

$u(x,y) = \frac{-1}{2} + e^{x-\frac{1}{4}y} + \frac{1}{2}e^{\frac{1}{2}y}\,$

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