# PDE1

$3u_{x}+4u_{y}-2u=1,u(x,0)=e^{x}\,$

Introduce the change of coordinates

$\xi =Ax+By\,$
$\eta =Cx+Dy\,$

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

${\frac {\partial u}{\partial x}}={\frac {\partial u}{\partial \xi }}{\frac {\partial \xi }{\partial x}}+{\frac {\partial u}{\partial \eta }}{\frac {\partial \eta }{\partial x}}=Au_{\xi }\ +Cu_{\eta }\,$,

${\frac {\partial u}{\partial y}}={\frac {\partial u}{\partial \xi }}{\frac {\partial \xi }{\partial y}}+{\frac {\partial u}{\partial \eta }}{\frac {\partial \eta }{\partial y}}=Bu_{\xi }+Du_{\eta }\,$.

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 $AD-BC\neq 0$. For example, let $C=1,D=-3/4$ so that the $u_{\eta }$ term goes away. Then let $A=1/3,B=0$ so that the $u_{\xi }$ coefficient equals unity. Then,

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

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

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

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

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

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

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

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

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.

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