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**leads** Show that if $\displaystyle u(x,t)$ is a solution of $\displaystyle u_{t}=u_{xx}$, then so is

$\displaystyle v(x,t)=t^{\frac{-1}{2}} e^{\frac{-x^{2}}{4t}}u(\frac{x}{t},-\frac{1}{t})$

I'm assuming that I should simply show $\displaystyle v_{t}=v_{xx}$ however I'm not sure how I should differentiate $\displaystyle u(\frac{x}{t},-\frac{1}{t})$ with respect to $\displaystyle t$.

??? $\displaystyle (u(\frac{x}{t},-\frac{1}{t}))'=t^{-4}u(\frac{x}{t},-\frac{1}{t}) $

Is this correct?