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Transform this equation: u_{{t}}=\nu u_{{xx}}+\lambda u_{x}+\mu u\, into the standard heat equation: v_{t}=v_{{xx}}\,

Assume a solution in the form u(x,t)=e^{{\alpha x+\beta t}}v(x,t)\,

Find its derivatives and plug it into the orginal DE:

u_{t}=\beta e^{{\alpha x+\beta t}}v+e^{{\alpha x+\beta t}}v_{t}\,

u_{x}=\alpha e^{{\alpha x+\beta t}}v+e^{{\alpha x+\beta t}}v_{x}\,

u_{{xx}}=\alpha ^{2}e^{{\alpha x+\beta t}}v+\alpha e^{{\alpha x+\beta t}}v_{x}+\alpha e^{{\alpha x+\beta t}}v_{x}+e^{{\alpha x+\beta t}}v_{{xx}}\,

\beta v+v_{t}=\nu (\alpha ^{2}v+2\alpha v_{x}+v_{{xx}})+\lambda (\alpha v+v_{x})+\mu v\,

Collect derivatives of v\, and find the appropriate values of \alpha \, and \beta \, to make the equation look like the heat equation.

v(\beta -\nu \alpha ^{2}-\lambda \alpha -\mu )+v_{t}-v_{x}(2\alpha \nu +\lambda )-v_{{xx}}(\nu )=0\,

\beta -\nu \alpha ^{2}-\lambda \alpha -\mu =0\,

2\alpha \nu +\lambda =0\,

\lambda =-2\alpha \nu \,

\alpha ={\frac  {-\lambda }{2\nu }}\,

\beta -\nu \left({\frac  {-\lambda }{2\nu }}\right)^{2}-\lambda \left({\frac  {-\lambda }{2\nu }}\right)-\mu =0\,

\beta -{\frac  {-\lambda ^{2}}{4\nu }}+{\frac  {2\lambda ^{2}}{4\nu }}-\mu =0\,

\beta =\mu -{\frac  {\lambda ^{2}}{4\nu }}\,

So the solution is

u(x,t)=e^{{{\frac  {-\lambda }{2\nu }}x+\left(\mu -{\frac  {\lambda ^{2}}{4\nu }}\right)t}}v(x,t)\,

Taking derivatives and plugging back in gives this equation:

v_{t}=\nu v_{{xx}}\,

Partial Differential Equations

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