Example 4. Solve the integro-differential equation

\[
\begin{gathered}
\varphi^{\prime \prime}(x)+\int_{0}^{x} e^{2(x-t)} \varphi^{\prime}(t) d t=e^{2 x} \\
\varphi(0)=\varphi^{\prime}(0)=0
\end{gathered}
\]
Solution. Let $\varphi(x) \equiv \Phi(p)$. By (11)

$$
\varphi^{\prime}(x) \risingdotseq p \Phi(p), \quad \varphi^{\prime \prime}(x) \risingdotseq p^{2} \Phi(p)
$$

Therefore, after applying the Laplace transform, equation (10) will take the form

$$
p^{2} \Phi(p)+\frac{p}{p-2} \Phi(p)=\frac{1}{p-2}
$$

or

$$
\Phi(p) \frac{p(p-1)^{2}}{p-2}=\frac{1}{p-2}
$$

From (12) we find

$$
\Phi(p)=\frac{1}{p(p-1)^{2}} \risingdotseq x e^{x}-e^{x}+1
$$

Therefore, the solution $\varphi(x)$ of the integro-differential equation (10), satisfying the initial conditions (11), is determined by the equality

$$
\varphi(x)=x e^{x}-e^{x}+1
$$

## Problems for Independent Solution

Solve the following integro-differential equations:

257. $\varphi^{\prime \prime}(x)+\int_{0}^{x} e^{2(x-t)} \varphi^{\prime}(t) d t=e^{2 x} ; \varphi(0)=0, \varphi^{\prime}(0)=1$.
258. $\varphi^{\prime}(x)-\varphi(x)+\int_{0}^{x}(x-t) \varphi^{\prime}(t) d t-\int_{0}^{x} \varphi(t) d t=x ; \varphi(0)=-1$.
259. $\varphi^{\prime \prime}(x)-2 \varphi^{\prime}(x)+\varphi(x)+2 \int_{0}^{x} \cos (x-t) \varphi^{\prime \prime}(t) d t+2 \int_{0}^{x} \sin (x-t) \varphi^{\prime}(t) d t=\cos x$; $\varphi(0)=\varphi^{\prime}(0)=0$.
260. $\varphi^{\prime \prime}(x)+2 \varphi^{\prime}(x)-2 \int_{0}^{x} \sin (x-t) \varphi^{\prime}(t) d t=\cos x ; \quad \varphi(0)=\varphi^{\prime}(0)=0$.
261. $\varphi^{\prime \prime}(x)+\varphi(x)+\int_{0}^{2} \operatorname{sh}(x-t) \varphi(t) d t+\int_{0}^{x} \operatorname{ch}(x-t) \varphi^{\prime}(t) d t=\operatorname{ch} x$; $\varphi(0)=\varphi^{\prime}(0)=0$.
262. $\varphi^{\prime \prime}(x)+\varphi(x)+\int_{0}^{x} \operatorname{sh}(x-t) \varphi(t) d t+\int_{0}^{x} \operatorname{ch}(x-t) \varphi^{\prime}(t) d t=\operatorname{ch} x ; \varphi(0)=-1$, $\varphi^{\prime}(0)=1$.

4 $^{\circ}$. Volterra Integral Equations with Limits $(x,+\infty)$. Integral equations of the form

$$
\varphi(x)=f(x)+\int_{x}^{+\infty} K(x-t) \varphi(t) d t
$$

arising in various problems of physics, can also be solved using the Laplace transform.

The following formula holds:

$$
\int_{x}^{+\infty} K(x-t) \varphi(t) d t \rightleftharpoons \overline{\mathscr{K}}(-p) \Phi(p)
$$

where

$$
\varphi(t) \risingdotseq \Phi(p), \quad \widetilde{\mathscr{K}}(-p)=\int_{0}^{+\infty} K(-x) e^{p x} d x
$$

Applying the Laplace transform to both sides of (13) and using formula (14), we get

$$
\Phi(p)=F(p)+\overline{\mathscr{X}}(-p) \Phi(p)
$$

or

$$
\Phi(p)=\frac{F(p)}{1-\widetilde{\mathscr{K}}(-p)} \quad(\widetilde{\mathscr{K}}(-p) \neq 1)
$$

The function

$$
\varphi(x)=\frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} \frac{F(p)}{1-\widetilde{\mathscr{K}}(-p)} e^{p x} d p
$$

is a particular solution of the integral equation (13). It should be emphasized that for the solution (15) to make sense, it is necessary that the regions of analyticity of $\overline{\mathscr{K}}(-p)$ and $F(p)$ overlap.