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Consider the partial differential equation:

$$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2} + a \frac{\partial u}{\partial x} + bu$$

for the function $u (x; t)$ where $a$ and $b$ are constants.

By using substitution of the form $u(x,t) = \exp(\alpha x+\beta t)v(x,t)$;

And suitable choice of constants alpha and beta, show that the PDE can be reduced to the heat equation

$$\frac{\partial v}{\partial t}=\frac{\partial^2v}{\partial x^2}.$$

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closed as off-topic by Behaviour, Claude Leibovici, RecklessReckoner, tetori, Carl Mummert Jul 15 at 10:03

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Just differentiate $v$ with respect to $x,x^2$ and $t$. Add the terms and you should see your reslut. Note that $v$ is determined by the substitution. Usually, you should show a bit more effort and tell us what you have tried so far. –  Quickbeam2k1 Feb 4 at 6:36

1 Answer 1

you could compare the equations $$\beta v+ \frac{\partial v}{\partial t} \\=\alpha^2v+\frac{\partial^2 v}{\partial x^2}+2\alpha\frac{\partial v}{\partial x}+a\alpha v+a\frac{\partial v}{\partial x}+bv$$ and the old one and choose the constant that fits your request.

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