The Problem
If you are given a nonlinear PDE
$$\frac{\partial c}{\partial t} = D\nabla^2c + \alpha c \, \, \,, \vec{r} \in \Omega, \,\, \, t\gt 0 \, \, \, \, (1)$$ where $D, \alpha$ are constants.
And then you are introduced a function defined by: $$\phi(\vec{r}, t) = c(\vec{r}, t)e^{-\alpha t} \, \, \, \, (2)$$ where $c$ satisfies $(1)$.
You are asked to derive the PDE for the function $\phi$
How does one solve this? My attempt is given below, but I have no proper idea of what I am doing.
My attempt
I find a way to express $c$ in terms of $\phi$: $$c = \phi e^{\alpha t} \, \, \, \, (3)$$ Then I just insert $(3)$ into $(1)$ to get: $$\frac{\partial(\phi e^{\alpha t})}{dt} = D\nabla^2\phi e^{\alpha t} + \alpha \phi e^{\alpha t} \, \, \, \, (4)$$
Is this the solution, or are there more steps to the problem? If someone could give me a guiding hand I would appreciate it very much.
Continuation after assitance from user Ninad Munshi
$$e^{\alpha t}\frac{\partial\phi}{\partial t}+ \alpha \phi e^{\alpha t} = D\nabla ^2 \phi e^{\alpha t} + \alpha \phi e^{at} $$ Dividing by $e^{\alpha t}$ $$\frac{\partial\phi}{\partial t}+ \alpha \phi = D\nabla ^2 \phi + \alpha \phi $$ Taking $- \alpha \phi$ on both sides $$\frac{\partial\phi}{\partial t} - D\nabla ^2 \phi = 0 $$ $$\frac{\partial\phi}{\partial t} - D\frac{\partial^2\phi}{\partial x^2} = 0 $$
Your new solution is right. You may recognize $\phi=c e^{-\alpha t}$ from the method of integrating factors for solving the ODE in $t$, $$ \frac{dc}{dt}(t) - \alpha c(t) = f(t).$$ Your exercise is essentially an application of this to the new setting of a certain PDE. It somehow worked out because $\nabla^2 \phi = (\nabla^2 c)e^{-\alpha t}$, by which I mean that this change of variables simplifies the equation from $(\partial_t - \alpha -D\nabla^2)c=0$ to the same equation (for $\phi$) but with $\alpha=0$. Since knowing $\phi$ tells you everything about $c$, it suffices to study the $\alpha=0$ case.
This is not what happens if $\alpha=\alpha(x)$ was a function of $x$; going through the same motions, if we set $\phi(x,t) = c(x,t)e^{-a(x)t}$, we get instead $$e^{\alpha(x)t}\partial_t \phi =(\partial_t-\alpha(x)) (\phi e^{\alpha(x)t}) = D\nabla^2(\phi e^{\alpha(x)t}) = e^{\alpha(x)t}D\left[\nabla^2 \phi + 2\nabla\alpha\cdot\nabla \phi + \phi \nabla^2 \alpha + \phi|\nabla \alpha|^2\right]$$ which does not lead to a simpler equation.