I have the heat equation $$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2}$$ where $0\leq t<\infty$ and $0\leq x\leq1$
The Boundary conditions are $u(0,t)=0$ and $u(1,t)=1$ with initial condition $u(x,0)=x+sin(\pi x)$
I assume the solution is of the form $u=u(x,t)=X(x)T(t)$ I find the respective derivatives of $u$ and substitute them into the original heat equation to get $$\frac{T'}{T}(t)=\frac{X''}{X} (x)=k$$
I have two Ode's $$T'=KT$$ and $$X''=KX$$
I now look at three different cases.
Case 1. K=0 My solutions say $X(x)=1$ and $T(t)=1$. Where do they come from?
For case 2. I have $K=-p^2<0$ The solution I have is $$X(x)=Acos(px)+Bsin(px)$$ and then $X(x)=sin(\pi nx)$and $T(t)=e^{-\pi^2n^2t}$ How do I get this solution?
For case 3. I have $K=p^2>0$. This gives me $X(x)=Ae^{px}+Be^{-px}$ whic gives me $X(x)=0$ and $T(t)=e^{p^2 t}$ Where do these come from?