Let $ s = xt^{-1/2} $ and look for the solution to the heat equation $ u_{t} = u_{xx} $ which is of the form: $ u(x,t) = t^{-\frac{1}{2}}f(s) $, which satisfies the condition $ \int_{-\infty}^{\infty} u(x,t) \; \mathrm{d}x = 1.$
This is what I've tried so far:
$ u = t^{-\frac{1}{2}}f(xt^{-\frac{1}{2}})$
$ u_{t} = -\frac{1}{2}t^{-\frac{3}{2}}f(xt^{-\frac{1}{2}}) + t^{-\frac{1}{2}}f'(xt^{-\frac{1}{2}})(-\frac{1}{2}xt^{-\frac{3}{2}})$
$ u_{t} = -\frac{1}{2}t^{-\frac{3}{2}} \left[ (f(s) + sf'(s) \right]$
$ u_{x} = t^{-\frac{1}{2}}f'(xt^{-\frac{1}{2}}) \cdot t^{-\frac{1}{2}} $
$ u_{xx} = t^{-\frac{1}{2}}f''(xt^{-\frac{1}{2}}) \cdot t^{-\frac{1}{2}} \cdot t^{-\frac{1}{2}} = t^{-\frac{3}{2}}f''(s)$
$ u_{t} = u_{xx}$
$ -\frac{1}{2}t^{-\frac{3}{2}} \left[ (f(s) + sf'(s) \right] = t^{-\frac{3}{2}}f''(s) $
$ f''(s) + \frac{s}{2}f'(s) + \frac{1}{2}f'(s)=0$
Basically, this is where I get stuck.
$$f^{′′}(s)+\frac{s}{2}f^′(s)+\frac{1}{2}f(s)=0$$ $$f^{''}(s)+{(\frac{s}{2} f(s))}^{'} = 0 $$ $$f^{'}(s)+\frac{s}{2}f(s) = A $$
$$ (f(s).\exp(\frac{s^2}{4}))^{'} = A\exp(\frac{s^2}{4}) $$ $$f(s).\exp(\frac{s^2}{4}) = f(0)+\int_{u=0}^s A\exp(\frac{u^2}{4}) \:du$$
you can integrate at any point you want , i chose the 0 but you can choose another one