This is a question on the lecture notes.
Basically we have the usual heat equation: $$\frac{\partial y}{\partial t}(x,t)=k^2\frac{\partial^2 y}{\partial^2 x}(x,t)+F(x,t)$$ We also have the usual trivial Dirichlet boundary conditions. Then the notes go on, saying that we make assumption about spatial dependence of solution can still be described by a Fourier Series, and because of the trivial Dirichlet BC we have the usual Fourier Sine Series. We then also have that the source term will as well be a Fourier Sine Series: $$\sum [y'_n(t)+(kn\pi)^2y_n(t)-F_n(t)]sin(n\pi x)=0$$ And then the notes said:
By using the standard orthogonality relations, we immediately see that: $$y'_n(t)+(kn\pi)^2y_n(t)-F_n(t)=0$$
How is this so? So far I have only known the usual integral orthogonality conditions. I assume that $y'_n(t)+(kn\pi)^2y_n(t)-F_n(t)$ and $sin(n\pi x)$ are both eigenfunctions, so they will be orthogonal, but that does not mean one of the terms is $0$?
If you have $\sum_n a_n \sin(n\pi x) = 0$ for all $x$ in your interval, then by multiplying by $\sin(m \pi x)$ and integrating over the interval you get that all $a_n = 0$. That's exactly what's happening here, with $a_n = y'_n(t) + (kn\pi)^2 y_n(t) - F_n(t)$.