Evans PDE book presents the following problem (page 87):
Write down an explicit formula for a solution of $$\left\{\begin{matrix} u_t-\Delta u+cu=f &\text{ in }\mathbb{R}^n\times(0,\infty) \\ u=g&\text{ on }\mathbb{R}^n\times\{0\} \end{matrix}\right.\tag{1}$$ where $c\in\mathbb{R}$.
In the text (page 51) was given a solution of
$$\left\{\begin{matrix} v_t-\Delta v=F &\text{ in }\mathbb{R}^n\times(0,\infty) \\ v=G&\text{ on }\mathbb{R}^n\times\{0\} \end{matrix}\right.\tag{2}$$
So, it's seems natural to try to reduce $(1)$ to $(2)$. This was done here (page 10).
My question (possibly elementary) is: If no one give us the answer, how can we discover that we should set $v(x,t)=u(x,t)e^{ct}$? What's the insight behind this choose?
Thanks.
Since the general solution of $u_t+cu=0$ is $u(x,t)=C(x)e^{ct}$ , therefore let $v(x,t)=u(x,t)e^{ct}$ can solely eliminating the term $cu$ of the PDE.