<s>[INST] Apply product and chain rule to differentiate with respect to one variable I have the following:
$$
u = \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}} \phi\left(\frac{x}{t}\right)
$$
I need to find $u_t$
This equation was found by solving the characteristic equation:
$$
\frac{dx}{xt} = \frac{dt}{t^2} = \frac{du}{-\left(\frac{1}{4}x^2 + \frac{1}{2}t\right)u}
$$
solving the equation from the first two yields:
$$
r = \frac{x}{t}
$$
solving from the second two yields:
$$
v = u\sqrt{t}e^{\frac{x^2}{4t}}
$$
we let $v = \phi(r)$
this gives us:
$$
u\sqrt{t}e^{\frac{x^2}{4t}} = \phi\left(\frac{x}{t}\right)
$$
that is where the equation for $u$ comes from.
Here is what I have:
\begin{align}
u_t & = \frac{-1}{2t^{\frac{3}{2}}}e^{\frac{-x^2}{4t}}\phi\left(\frac{x}{t}\right) + \frac{x^2}{4t^2}\frac{1}{\sqrt{t}}e^\frac{-x^2}{4t}\phi\left(\frac{x}{t}\right) + \left( \frac{-x}{t^2} \right) \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi'\left(\frac{x}{t}\right) \\
&  = \frac{-u}{2t} + \frac{x^2u}{4t^2} - \left( \frac{x}{t^2} \right) \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi'\left(\frac{x}{t}\right) \\ 
\end{align}
However the solution given has:
$$
u_t = \frac{-u}{2t} + \frac{x^2u}{4t^2} - \phi'\left( \frac{x}{t} \right) \left(\frac{xu}{t^2}\right)
$$
I can't figure out what I am doing wrong before I look at $u_{xx}$ (I already tried that and again, am making the same mistake).
 [/INST] Your mistake is in the second term. It should be
$$\frac{x^2}{4t^{2}}\frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi\left(\frac{x}{t}\right) = \frac{x^2}{4t^{2}}u$$
You just forgot the chain rule.
</s>