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