Solving PDE using Separation of Variables with Dirichlet boundary conditions

Question

Solve the following PDE, $$ u_t(x,t)=ku_{xx}(x,t)-bu(x,t)$$ where $b>0$, with boundary conditions $$u(0,t)=u(c,t)=0 $$

My attempt

Assume $u(x,t)=X(x)T(t)$ and plugging into the diffirential equation,

$$X(x)T'(t)=kX''(x)T(t)-bX(x)T(t)$$

$$\dfrac{T'(t)}{T(t)}=k\dfrac{X''(x)}{X(x)}-b=-\lambda$$

Then solving $X(x)$ gives the ODE

$$kX''(x)+(\lambda-b)X(x)=0$$

with boundary conditions,

$$X(0)=X(c)=0$$

How to go from here?

Edit

Setting up the characteristic equation, $$r^2+\dfrac{\lambda-b}{k}=0$$ gives, $$r=i\mkern1mu\sqrt{\dfrac{\lambda-b}{k}}$$ which gives a general solution, $$X(x)= c_1\cos\bigg(\sqrt{\dfrac{\lambda-b}{k}}x\bigg)+c_2\sin\bigg(\sqrt{\dfrac{\lambda-b}{k}}x\bigg)$$ And from here her I don't know

Answer 1

You can rewrite the equation as $$ u_t(x,t)+bu(x,t)=ku_{xx}(x,t) $$ Multiplying by $e^{bt}$ gives the more standard form: $$ (e^{bt}u)_{t}=(e^{bt}u)_{xx} $$ Therefore $v(x,t)=e^{bt}u(x,t)$ satisfies $$ v_{t}=kv_{xx} \\ v(0,t)=0=v(c,t). $$ Using separation of variables gives a solution $$ v(x,t) = \sum_{n=1}^{\infty}A_ne^{-n^2\pi^2kt/c^2}\sin(n\pi x/c). $$ The constants $A_n$ are determined through orthogonality from the initial data $v(x,0)$. Then $u(x,t)=e^{-bt}v(x,t)$.

Answer 2

There are a discrete infinity of values of lambda that correspond to the number of half sine waves that can fit in the interval. The general solution is a sum over all of them. The coefficients are found from the initial condition using orthogonalty of the sines.