Let's say I have the following boundary value problem: $$ \partial_t u(x,t)=\partial_{xx}u(x,t)-u(x,t)+sin(\pi x), \quad x\in(0,1),\quad t>0 $$ $$ u(0,t)=0, \quad t>0 $$ $$ u_x(1,t)=0, \quad t>0 $$ $$ u(x,t)=a, \quad x\in(0,1) $$
To find the Eigenvalues and Eigenvectors I found the equation for this to be: $$ \phi''(x)+\lambda\phi=0, \quad x\in (0,1) $$ $$ \phi(0)=\phi_x(1)=0 $$
However: I couldn't figure out why I have to use this equation. And how do I find the equation for other boundary-value problems? Is there a general way for this?
First we solve the homogeneous problem using separation of variables; Let us suggest a solution of the form - $u(x,t)=X(x)T(t)$. Plugging this into the PDE we get:
$$T'X-TX''+TX=0$$ $$\frac{X''}{X}=\frac{T'}{T}+1$$
Since both sides of the equation depend strictly on two independent variables, we conclude that they are both equal to some constant $-\lambda$ (the minus sign is for convenience). We thus get an equation for the term that depends on $x$: $$X''+\lambda X=0$$ Which is the equation you were given. From the given boundary conditions we know that - $$X(0)=X_x(1)=0$$ (since the result needs to be independent of the function $T$)
This explains why you need to use this equation. You can now solve the equation in order to find the part of the solution that depends on $X$. You will get a set of eigenfunctions that span the space of solutions. You can then use the temporal initial condition and the in-homogeneous part of the equation in order to find the complete solution.