Passage from the Cauchy problem to the integral equation 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?
 A: 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
$$y(x) = \frac{\sin \rho x}{\rho} + \int_0^x \frac{\sin \rho (x-t)}{\rho} \left( q(t)y(t) + \int_0^t M(t-s)y(s) ds\right) dt.$$
