Consider the following PDE:
$$\frac{\partial }{\partial t}u = \alpha^2 \frac{\partial^2 }{\partial x^2}u -bu, \ \ b>0, 0<x<L$$
The problem asks to "set the time derivative in the PDE to zero, solve for the steady state solution".
By separation of variables, let $u(t,x)=T(t)X(x)$ and plug in PDE. We get
$$0 = \alpha^2 T \frac{d^2 X }{d x^2}u-bTX \Longrightarrow \frac{d^2 X }{d x^2}-\frac{b}{\alpha^2}X=0$$
We get $$X(x) = A\sin\left(\sqrt{-\frac{b}{\alpha^2}}x\right)+B\cos\left(\sqrt{-\frac{b}{\alpha^2}}x\right)$$
I am a bit confused about the answer I get here since it is a complex number. But the answer should be
$$u(t,x) = e^{-bt} u_0(t,x)$$ where $u_0(t,x)$ solves the heat conduction in a rod (without loss). I am not sure where I made error.
The solution to the differential equation is sum of exponential functions.
$$\frac{\partial^2 f}{\partial t^2} - b^2f(t) = 0$$
$$f(x) = k_1 \exp(-bt) +k_2 \exp(bt)$$
If you had opposite sign of $bTX$ then it would be complex conjugate roots and sine-cosine linear combination:
$$\frac{\partial^2 f}{\partial t^2} + b^2f(t) = 0$$
giving
$$f(x) = k_1 \sin(bt) +k_2 \cos(bt)$$