Making this question, so that I can understand thoroughly how such a type of exercise gets solved, because our lectures have only been theoritical regarding theorems and their proofs, so our problem solving experience is non-existent in this moment.
Exercise :
Find the Integral Equation which is equivalent to the initial value problem :
$$y'' + μ^2y = g(t,y)$$
$$y(0) = y_0$$ $$y'(0) = z_0$$
where $g(t,y)$ is continuous in a field $D$ that contains $(0,y_0)$ and $μ>0$.
Note :
I apologize for not providing any attempt, but as I said, our experience solving problems is non-existent at this moment and me and my co-students are trying to start understanding some exercises. Any help or thorough solution would be much appreciated !
As nearly hinted at in the formulation of the initial values, you can also use the first order system $y'=z$, $z'=g(t,y)-μ^2y$ and use the standard integral equation for first order systems $$ \begin{pmatrix}y'(t)\\z'(t)\end{pmatrix} = \begin{pmatrix}y_0\\z_0\end{pmatrix} + \int_0^t \begin{pmatrix}z(s)\\g(s,y(s))-μ^2y(s)\end{pmatrix}\,ds. $$
Or consider $w(t)=e^{iμt}(μy(t)+iy'(t))$ so that $$ w'(t)=e^{iμt}\Bigl(μy'(t)+iy''(t)+iμ\bigl(μy(t)+iy'(t)\bigr)\Bigr)=ie^{iμt}g(t,y(t)) $$ and the equivalent integral equation is $$ μy(t)+iy'(t) = \bigl(μy(0)+iy'(0)\bigr)e^{-iμt} + i\int_0^t e^{iμ(s-t)}g(s,y(s))\,ds $$ where the real component is an integral equation for $y$, $$ y(t)=y(0)\cos(μt)+y'(0)\frac{\sin(μt)}{μ}+\int_0^t\frac{\sin(μ(t-s))}{μ}g(s,y(s))\,ds $$
This is just a systematic way to write down the solution of $$y''(t)+μ^2y(t)=f(t)$$ where $f(t)$ is replaced by $g(t,y(t))$ in the solution formula. In the same spirit you could also solve $$y''(t)=f(t)$$ and replace $f(t)=g(t,y(t))-μ^2y(t)$ in the solution formula \begin{align} y(t)&=y(0)+y'(0)t+\int_0^t(t-s)f(s)\,ds \\ &=y(0)+y'(0)t+\int_0^t(t-s)\bigl(g(s,y(s))-μ^2y(s)\bigr)\,ds \end{align}
In general, the integral form is a theoretical device. It maps continuous functions to differentiable functions. As the projection from $C^1$ to $C^0$ is a compact operator, this turns the whole fixed-point mapping into a compact mapping from $C^0$ to $C^0$ from which follows the existence of fixed points and thus solutions, or in the context of Sturm-Liouville boundary value problems, the structure of the spectrum. The more information you have on the right side $g$, the more properties of the fixed-point you can prove.