PDE1

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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}}\,

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