# Find $u(x,t)$ which is defined in terms of another *unknown* function $\phi$ as $t^{-1/2} \phi \left(\frac{x}{\sqrt{t}}\right)$

So I have:

$u(x,t) = t^{-1/2} \phi (\xi) \,\,\, where \,\, \xi=\frac{x}{\sqrt{t}}$

and I know that:

$\frac{du}{dt} = 1/2\frac{d^2u}{dx^2}$

And

$\int_{-\infty }^{\infty } u(x,t) \, dx = 1$

And for $\phi(\xi)$ I have:

$\phi′′+\xi\phi′+\phi=0$

How do I find $u(x,t)$ ?

The only "shortcut" I'm given is that I can assume without proof that:

$\int_{-\infty }^{\infty } e^{-\xi^2/2} d\xi = \sqrt{2 \pi }$

I know the answer but I don't know how to get it:

$u(x,t) = \frac{e^{-\frac{X^2}{2 t}}}{\sqrt{2 \pi }}$

I can guess that it will be solved using the Integrating Factor Method.

Any hints or directions will be greatly appreciated.

Thanks

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(1) You should be comfortable transforming between the PDE for $u(x,t)$ and ODE for $\phi(\xi)$ by using the chain rule. Don't think of them as two separate equations - the equation for $\phi$ is just a reformulation of the equation for $u$ under the assumption that $u$ has a particular shape.
(2) If we could solve for $\phi$ then we immediately recover $u$ from $u = \frac{1}{\sqrt{t}}\phi$.
(3) We can't use the integrating factor method straight away to solve for $\phi$ since there is a second derivative. However, we can reduce the order of this equation by noticing that $$\xi \phi' + \phi = (\xi\phi)'.$$