Can we determine the solution to $u_t+bu=u_{xx}$ given the solution to $u_t=u_{xx}$?

Question

If the solution to $$u_t=u_{xx}, \ \ u(x,0)=\phi(x)$$ is $$u(x,t)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty} e^{-(x-y)^2/4t}\phi(y) \ dy,$$ what is the solution to $$u_t+bu=u_{xx}, \ \ u(x,0)=\phi(x) \tag{1}$$ where $b>0$ is a constant.

This problem reminds me of finding homogeneous and inhomogeneous solutions to simple ODEs. I though that $$u_t-u_{xx}=0,$$ could represent the homogeneous solution, $u_H(x,t)$, meaning we require a particular solution, $u_P(x,t)$, to $$u_t-u_{xx}=bu.$$ Hence the solution to $(1)$ is $u(x,t)=u_H(x,t)+u_P(x,t).$ I hint would be very helpful.

Answer 1

Let $\mu(t)$ be an "integrating factor" such that

$$ \mu u_t + b\mu u = \big[\mu u\big]_t $$

Then the original PDE can be simplified to

$$ v_t = v_{xx} $$

where $v(x,t) = \mu(t) u(x,t)$

The first equation simplifies to $\mu' = b\mu$, or $\mu(t) = e^{bt}$

This is somewhat analogous to the first-order ODE

$$ y'(t) + by(t) = f(t) $$

Answer 2

I guess the second equation is equivalent to $$ v_t = v_{xx} $$ with $v(x,0)=\phi(x)$, where $v(x,t) = e^{bt}u(x,t)$.