How can I show that this function is a solution to $y''(t) + 4y(t) = f(t)$? I was reading this document about Green's functions and on pages $242-43$ the author talks about a series representation for Green's functions. The author then gives an example of this representation to show that a solution to the differential equation
$$
y''(t) + 4y(t) = f(t)\tag{1}
$$
with $0<t<1$ and $y(0) = 0 =y(1)$ is given by the function
$$
y(t) = \int_0^1 \underbrace{\left(2 \sum_{n\ge1} \frac{\sin(n\pi t)\sin(n \pi \zeta)}{4- (n\pi)^2} \right)}_{\color{blue}{G(t, \zeta)}}f(\zeta) \ d\zeta \tag{2}
$$
The proof as to why this series representation is valid (as shown in the document) involves eigenfunctions of a differential operator and other concepts I'm not very familiar with. Despite this, I was wondering if I could "backtrack" the steps and prove that the explicit solution in equation $(2)$ satisfies the differential equation $(1)$ by direct substitution, thereby showing that this solution is valid without going into subjects I don't understand quite well.

My attempt
If I substitute directly, using the Leibniz rule for integration I get that
\begin{align}
y'' + 4y &=\int_0^1 \frac{\partial^2}{\partial t^2}\left(2 \sum_{n\ge1} \frac{\sin(n\pi t)\sin(n \pi \zeta)}{4- (n\pi)^2} \right)f(\zeta) \ d\zeta  + 4 \int_0^1 \left(2 \sum_{n\ge1} \frac{\sin(n\pi t)\sin(n \pi \zeta)}{4- (n\pi)^2} \right)f(\zeta) \ d\zeta\\
& =\int_0^1 \sum_{n\ge1} \left(8 - 2 n^2 \pi^2 \right)\frac{\sin(n\pi t)\sin(n \pi \zeta)}{4- (n\pi)^2} f(\zeta) \ d \zeta \\
& =\int_0^1 \sum_{n\ge1} 2\sin(n\pi t)\sin(n \pi \zeta) f(\zeta) \ d \zeta \\
& = \sum_{n\ge1} 2\sin(n\pi t) \int_0^1\sin(n \pi \zeta) f(\zeta) \ d \zeta
\end{align}
And this is where I got stuck since I don't know how I could show that this last part is equal to $f(t)$. Can anyone tell me if it's possible to directly show that equation $(2)$ satisfies $(1)$ in a similar manner to what I tried? Or alternatively, can someone give me a hint as to how I could finish my solution? Thank you!
 A: This or similar examples are usually used to introduce the idea of using the eigenbasis to solve general Sturm-Liouville problems.
The identity in question is the sine series identity with the original function, a particular case of the Fourier series identity
$$
f(t)=\frac{a_0}2+\sum_{k>0}a_k\cos(k\pi t)+b_k\sin(k\pi t),
\\~~~
a_k=\int_{-1}^1\cos(k\pi s)f(s)\,ds,
~~~
b_k=\int_{-1}^1\sin(k\pi s)f(s)\,ds.
$$
For that imagine $y$ and $f$ continued as $2$-periodic function with an odd symmetry at the origin. Then the even cosine components get coefficients $a_k=0$, and only the sine part of the Fourier series remains, with $b_k=2\int_0^1\sin(k\pi s)f(s)\,ds$.
The equations $y''(t)+4y(t)=\sin(k\pi t)$ for the sine series "atoms" have well-known solutions, finite linear combinations of the right side have the associated linear combinations of the "atomic" solutions as solution. In question is now the transition to the sine series expansion of general right sides.
A halfway elementary discussion of Fourier series expansions can be found in Carl Offner: "A little harmonic analysis" where the Dirac delta is identified as unit of the convolution operator, approximations of delta are defined in this context and appropriate "mollifications" of the Fourier series with their delta-comb like kernels are discussed.
