I try understand one paper about inverse spectral problem for $$-y''+q(x)y + \int_0^x M(x-t)y(t)dt = \lambda y, \quad 0 < x < \pi, \quad y(0)=y(\pi)=0$$ There is statement that Cauchy problem of this equation with initial values $y(0)=0, y'(0) = 1$ is equivalent to the integral equation: $$y(x) = \frac{\sin \rho x}{\rho} + \int_0^x \frac{\sin \rho (x-t)}{\rho} \left( q(t)y(t) + \int_0^t M(t-s)y(s) ds\right) dt,$$ where $\rho = \sqrt \lambda$.
First question: why it is true? Second: where did it come from?
I found solution and I want try to write it.
We will use variation of constants method. Consider differential equation $$y'' + \lambda y = qy + \int_0^x M(x-t)y(t)dt.$$ At first we find solution for homogenious part: $$y'' + \lambda y = 0.$$ Common solution is $y= C_1 \cos \rho x + C_2 \sin \rho x = C_1 y_1 + C_2 y_2$, where $\rho = \sqrt \lambda$.
Okey, now let $C_1 = C_1(x), C_2 = C_2(x)$ - unknown functions. By variation of constants we should solve system: $$\begin{cases} C_1' y_1 + C_2'y_2 = 0\\ C_1' y_1' + C_2'y_2' = q(x)y + \int_0^x M(x-t)y(t)dt \end{cases}$$ or $$\begin{cases} C_1' \cos \rho x + C_2' \sin \rho x= 0\\ -C_1' \rho \sin \rho x + C_2'\rho \cos \rho x = q(x)y + \int_0^x M(x-t)y(t)dt. \end{cases}$$ Then \begin{align*} C_1' &= -\frac{C_2' \sin\rho x}{\cos \rho x} \Rightarrow \frac{C_2' \sin^2 \rho x}{\cos\rho x} \rho + C_2' \rho \cos \rho x = qy + \int_0^x M(x-t) y(t) dt \Rightarrow \\ &\Rightarrow C_2' = \frac{\cos \rho x}{\rho} \left(qy + \int_0^x M(x-t) y(t) dt\right) \Rightarrow\\ &\Rightarrow C_2(x) = \int_0^x \frac{\cos \rho t}{\rho} \left(qy + \int_0^t M(t-s) y(s) ds\right) dt + C_2 \end{align*} Here $C_2$ - arbitrary constant. Next, we get easily $$C_1(x) = -\int_0^x \frac{\sin \rho t}{\rho} \left(qy + \int_0^t M(t-s) y(s) ds\right) dt + C_1$$ and \begin{align*}y &= C_1(x)\cos\rho x + C_2(x) \sin \rho x = \ldots =\\ &= \int_0^x \frac{\sin \rho(x-t)}{\rho} \left(qy + \int_0^t M(t-s) y(s) ds\right) dt + C_1 \cos\rho x + C_2 \sin \rho x \end{align*} It remains determine constants from initial data: \begin{align*} &y(0) = C_1 = 0\\ &y'(0) = -C_1 \rho\sin\rho x + C_2 \rho\cos\rho x |_{x=0} = C_2 \rho = 1 \Rightarrow C_2 = \frac{1}{\rho}. \end{align*} Well, it is the end - we get that $$y(x) = \frac{\sin \rho x}{\rho} + \int_0^x \frac{\sin \rho (x-t)}{\rho} \left( q(t)y(t) + \int_0^t M(t-s)y(s) ds\right) dt.$$