Nondimensionalising and solving the heat equation with forcing term I am given an equation of the form : 
$$\frac{\delta T}{\delta t^*} = -\lambda(T-T_{\infty}) + K \frac{\delta^2T}{\delta x^{*2}}$$ 
over $0 < x* < \infty$ , t*>0
With initial condition: 
$$T(x^*,0) = T_{\infty} + T_0\left[1-\frac{x^*}{L^2}\right] , \space \space 0 \le x^* \le L$$
and $$T(x^*,0) = T_\infty , \space \space x^*>L$$
and insulating boundary conditions, ie $$\frac{\delta T}{\delta x^*}(0,t)=\frac{\delta T}{\delta x^*}(L,t)=0$$
I need to nondimensionalise the equation using the scaled variables:
$$t = \frac{Kt^*}{L^2}, x = \frac{x^*}{L}, \theta = \frac{(T-T_\infty)}{T_0}$$
My attempt at nondimensionalising results in:
$$\frac{\delta \theta}{\delta t} = -\gamma * \theta + \frac{\delta ^2 \theta}{\delta x^2}$$
where $ \gamma =\frac{\lambda L^2}{K}$
and initial condition becomes:
$$\theta (x,0) = 1-x^2 , \; \text{for} \; 0 \le x \le 1$$
$$\theta (x,0) = 0 ,\; \text{for}\; x > 1.$$
Is this correct? I have had trouble in the past nondimensionalising so I am not sure. 
We are then told that the internal heat is given by $$Q(t)=\int_{0}^{\infty}  \theta(x,t)\,dx$$
and to show that $$Q=Q_0e^{-\gamma t}.$$
If my nondimensionalised equation is correct, how do I then go about solving it? I assume that is what the second part requires. The forcing term seems in vary with both $x$ and $y$ which I do not know how to solve. (even if it is in only $x$ or $t$ I'm still uncertain how to solve it). Any help would be appreciated.
 A: Using successive (multivariable) chain rules, we find
$$ \frac{\partial \theta}{\partial t} = \frac{d\theta}{dT} \frac{\partial T}{\partial t} = \frac{d\theta}{dT}\frac{\partial T}{\partial t^*}\frac{dt^*}{dt} = \frac{L^2}{T_0K}\frac{\partial T}{\partial t^*} $$
Similarly
$$ \frac{\partial \theta}{\partial x} = \frac{d\theta}{dT}\frac{\partial T}{\partial x^*}\frac{dx^*}{dx} = \frac{L}{T_0}\frac{\partial T}{\partial x^*} $$
$$ \frac{\partial^2\theta}{\partial x^2} = \frac{L}{T_0}\frac{\partial }{\partial x}\left(\frac{\partial T}{\partial x^*}\right) = \frac{L}{T_0}\frac{\partial T}{\partial x^{*2}}\frac{dx^*}{dx} = \frac{L^2}{T_0}\frac{\partial T}{\partial x^{*2}} $$
Substituting in
$$ \frac{T_0K}{L^2} \frac{\partial\theta}{\partial t} = \frac{T_0K}{L^2}\frac{\partial \theta}{\partial x^2} -\lambda T_0\theta $$
$$ \frac{\partial \theta}{\partial t} = \frac{\partial^2 \theta}{\partial x^2} - \frac{\lambda L^2}{K} \theta $$
which verifies your result. The boundary conditions are also easy to verify as well.

You do not need to solve the remaining PDE. We can find an ODE for $Q(t)$ by differentiating under the integral sign
\begin{align} 
Q'(t) &= \int_0^\infty \frac{\partial \theta}{\partial t}(x,t) dx \\
&= \int_0^\infty\left[\frac{\partial^2 \theta}{\partial x^2}(x,t) - \gamma \theta(x,t)\right]\ dx \\
&= \frac{\partial \theta}{\partial x}(x,t)\Bigg|_{x=0}^{x=\infty} - \gamma \int_0^\infty \theta(x,t)\ dx \\
&= -\gamma Q 
\end{align}
which has a solution as given. You may also find that
$$ Q_0 = Q(0) = \int_0^\infty \theta(x,0)\ dx = \int_0^1 (1-x^2)\ dx = \frac23 $$
