Show a PDE satisfies a ODE Suppose that $u(x,t)$ satisfies the partial differential equation $${du\over dt} = \frac 12 {d^2u\over dx^2}$$ for all $t>0$ and $x\in R$, and also that has functional form $$u(x,t) = t^\alpha\omega(\theta)$$ where $$\theta=\frac x{\sqrt{t}}$$ How do I show that $\omega(\theta)$ must satisfy the following ordinary differential equation $$\omega'' + (\theta\omega)' = (1+2\alpha)\omega,$$ for all $\theta\in R$
I have tried to rearrange the functional form in terms of $\omega$, and diffentiate w.r.t. $t$, but not arriving at intended destination
 A: $$
\begin{align}
    \frac{\partial u}{\partial t}&=\frac{\partial}{\partial t}(t^\alpha w(\theta))\\
        &=\alpha t^{\alpha-1}w(\theta)+t^\alpha\frac{\partial \theta}{\partial t}w'(\theta)\\
        &=\alpha t^{\alpha-1}w(\theta)+t^\alpha\left(-\frac{1}{2t}\theta\right)w'(\theta)\\
        &=t^{\alpha-1}(\alpha\, w(\theta)-\theta/2\;w'(\theta))\\[10pt]
    \frac{\partial u}{\partial x}&=t^\alpha\frac{\partial \theta}{\partial x}w'(\theta)\\
        &=t^{\alpha-1/2}w'(\theta)\\[10pt]
    \frac{\partial^2 u}{\partial x^2}&=t^{\alpha-1}w''(\theta)
\end{align}
$$
Put it all together, notice that for $t\not=0$ we can simplify by $t^{\alpha-1}/2$ and get
$$
    2\alpha\,w(\theta)=w''(\theta)+\theta\,w'(\theta).
$$
Using the fact that $(\theta\,w)'=w+\theta\,w'$, we finally get
$$
    w''+(\theta\,w)'=(1+2\alpha)w.
$$
A: Hint: all the ordinary derivatives you write are, in reality, partial derivatives. You have
$$2\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}, $$
with
$$u(x,t)=t^{\alpha}\omega(\theta(x,t)) $$
and $\theta(x,t)=\frac{x}{\sqrt{t}}.$ Then your PDE is equivalent to
$$2\alpha t^{\alpha-1}\omega(\theta(x,t))+2t^{\alpha}\frac{\partial \omega(\theta(x,t))}{\partial t}=t^{\alpha}\frac{\partial^2 \omega(\theta(x,t))}{\partial x^2}. $$
All you need now is to apply the chain rule on the partial derivatives applied to the composite function $\omega(\theta(x,t))$. Can you do this?
