Variation of parameters: Consider IBVP \begin{align} u_t − u_{xx} = f(x, t) \qquad & \text{on } \Omega = (0, \pi) \times \Bbb R^+\\ u(x, 0) = \varphi(x) \qquad & \text{on } (0, \pi)\\ u(0, t) = h_0(t) \qquad & \text{on } \Bbb R^+\\ u(\pi, t) = h_1(t) \qquad & \text{on } \Bbb R^+ \end{align} Find $v(x, t) = a(t) + b(t)x$ which satisfies the two boundary conditions and write down the problem solved by $w = u − v$ (for the sake of clarity, define new names for complicated expressions).
So I believe the idea is that we want $w_t - w_{xx} = 0$ so it is reduced to what I have learned. We want $w_t - w_{xx} = u_t - u_{xx} - v_t = 0$. (Since we can get $v_{xx}=0$.) This means $v_t = f(x,t)$. However $v_t = a' + b'x$ has highest power to $x$ of $1$. What if $f$ is not of the form $a' + b'x$? Am I doing in the wrong way?