rewrite $\frac{\partial \theta}{\partial t}=k \frac{\partial^2 \theta}{\partial x^2}$ as a DE with two new variables $q_1$ and $q_2$

I am given the differential equation: $$\frac{\partial \theta}{\partial t}=k \frac{\partial^2 \theta}{\partial x^2}$$ Use the change of variables $q_1(x,t) = \frac{x^2}{kt}$ and $q_2 (x,t)=\frac{\theta(x,t)\sqrt{kt}}{\theta_0}$ to rewrite the differential equation in terms of $q_1$, and $q_2$.

I asked a similar question to this a couple of days ago but with the help of my professor I got a bit further and am taking a different approach to the problem now and so I am posting this question.

Now $$\frac{1}{\sqrt{kt}} = \frac{\sqrt{q_1}}{x}$$ $$\theta = \frac{q_2 \theta_0}{\sqrt{kt}}=\frac{q_2 \sqrt{q_1}}{x}\theta_0=\theta(q_1,q_2)$$

So up till here everything is fine but I can't seem to get the next step. I have to rewrite the original differential equation in terms of $q_1,q_2$, then assume that $q_2=f(q_1)$ and show that $f$ satisfies the ODE $$4q_1 \frac{\partial ^2f}{\partial q_1^2}+(q_1+2) \frac{\partial f}{\partial q_1}+\frac{f}{2}=0$$ Whenever I try to rewrite the original differential equation I end up with enormously complicated expressions that are impossible to simplify and dont lead to the given differential equation for $f$ at all. I am sure there is some method or way to solve this but I really can't find. If anyone could help me I would be very grateful because I have been stuck on this for days. It is also weird to me because the other homework questions I got I solved fairly quickly after some messing about. Thanks in advance!

Let $q_1=\dfrac{x^2}{kt},q_2=\dfrac{\theta\sqrt{kt}}{\theta_0}$ thus we have:$$\frac{\partial q_1}{\partial x}=\frac{2x}{kt}\\\frac{\partial q_1}{\partial t}=-\frac{x^2}{kt^2}\\\frac{\partial q_2}{\partial x}=\frac{\sqrt{kt}}{\theta_0}\frac{\partial\theta}{\partial x}\\\frac{\partial q_2}{\partial t}=\frac{\sqrt{k}}{2\theta_0\sqrt{t}}\theta_0+\frac{\sqrt{kt}}{\theta_0}\frac{\partial\theta}{\partial t}$$hence: $$\frac{\partial\theta}{\partial t}=\frac{\theta_0}{\sqrt{kt}}\left(\frac{\partial q_2}{\partial t}-\frac{\sqrt{k}}{2\theta_0\sqrt{t}}\theta\right)\\\frac{\partial\theta}{\partial x}=\frac{\theta_0}{\sqrt{kt}}\frac{\partial q_2}{\partial x}\\\frac{\partial^2\theta}{\partial x^2}=\frac{\theta_0}{\sqrt{kt}}\frac{\partial^2q_2}{\partial x^2}$$Hence substituting into our original equation we have:$$\frac{\theta_0}{\sqrt{kt}}\left(\frac{\partial q_2}{\partial t}-\frac{\theta\sqrt{k}}{2\theta_0\sqrt{t}}\right)=\frac{k\theta_0}{\sqrt{kt}}\frac{\partial^2q_2}{\partial x^2}\\\frac{\partial q_2}{\partial t}-\frac{\sqrt{k}}{2\theta_0\sqrt{t}}\theta=k\frac{\partial^2 q_2}{\partial x^2}$$Now notice we have $$\theta=\frac{\theta_0}{\sqrt{kt}}q_2\\\frac{\sqrt{k}}{2\theta_0\sqrt{t}}\theta=\frac1{2t}q_2$$giving us:$$\frac{\partial q_2}{\partial t}-\frac1{2t}q_2=k\frac{\partial^2 q_2}{\partial x^2}$$

Do you follow?

${\large% {\partial\theta \over \partial t} = k\, {\partial^{2}\theta \over \partial x^{2}}}$

$\large\tt\mbox{Scaling:}\$ $\large\theta \to \alpha\,\theta\,,\quad x \to \beta\,x\,,\quad t \to \gamma\,t$

$$\Longrightarrow\ {\beta^{2} \over \gamma}\,{\partial\theta \over \partial t} = k\, {\partial^{2}\theta \over \partial x^{2}}$$ The equation doesn't change its form whenever $\beta = \sqrt{\vphantom{\large a}\gamma\,}\quad$ or/and $\quad{\beta x \over x} = \sqrt{\gamma t \over t\,}\quad$ or/and $\quad{\beta x \over \sqrt{\gamma t\,}} = {x \over \sqrt{\vphantom{\large A}t\,}}$. That means that the change $y = {x \over \sqrt{\vphantom{\large A}t\,}}$ should "simplify" the equation:

\begin{align} {\partial \over \partial t} &= {\partial y \over \partial t}\,{{\rm d} \over {\rm d}y} = -\,{x \over 2\,t^{3/2}}\,{{\rm d} \over {\rm d}y} = -\,{y \over 2\,t}\,{{\rm d} \over {\rm d}y} \\[3mm] {\partial \over \partial x} &= {\partial y \over \partial x}\,{{\rm d} \over {\rm d}y} = {1 \over t^{1/2}}\,{{\rm d} \over {\rm d}y} \quad\Longrightarrow\quad {\partial^{2} \over \partial x^{2}} = {1 \over t}\,{{\rm d}^{2} \over {\rm d}y^{2}} \end{align}

The "new" equation is given by: $$-\,{y \over 2t\,}{{\rm d}\theta \over {\rm d}y} = k\,{1 \over t}\,{{\rm d}^{2}\theta \over {\rm d}y^{2}} \quad\Longrightarrow\quad k\,{{\rm d}^{2}\theta \over {\rm d}y^{2}} + {y \over 2\,}{{\rm d}\theta \over {\rm d}y} = 0 \quad\Longrightarrow\quad {{\rm d} \over {\rm d}y}\left(% {\rm e}^{y^{2}/2k^{2}}\,{{\rm d}\theta \over {\rm d}y} \right) = 0$$

Then, $${{\rm d}\theta \over {\rm d}y} = A\,{\rm e}^{-y^{2}/2k^{2}}\,, \qquad (~A\quad \mbox{is independent of}\quad y~).$$

$$\begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \theta\left(y\right) \color{#000000}{=} \theta\left(y_{0}\right) + A\int_{y_{0}}^{y}{\rm e}^{-y\,'^{2}/2k^{2}}\,{\rm d}y'\quad } \\ \\ \hline \end{array}$$