Example. Solve the integral equation

$$
\int_{0}^{x} e^{x-t} \varphi(t) d t=x
$$
Solution. Applying the Laplace transform to both sides of (3), we get

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

from which

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

The function $\varphi(x)=1-x$ is the solution to equation (3).

## Problems for Independent Solution

Solve the integral equations:

284. $\int_{0}^{x} \cos (x-t) \varphi(t) d t=\sin x$.
285. $\int_{0}^{x} e^{x-t} \varphi(t) d t=\operatorname{sh} x$.
286. $\int_{0}^{x}(x-t)^{1 / 2} \varphi(t) d t=x^{5 / 2}$.
287. $\int_{0}^{x} e^{2(x-t)} \varphi(t) d t=\sin x$.
288. $\int_{0}^{x} e^{x-t} \varphi(t) d t=x^{2}$.
289. $\int_{0}^{x} \cos (x-t) \varphi(t) d t=x \sin x$.
290. $\int_{0}^{x} \operatorname{sh}(x-t) \varphi(t) d t=x^{3} e^{-x}$.
291. $\int_{0}^{x} J_{0}(x-t) \varphi(t) d t=\sin x$.
292. $\int_{0}^{x} \operatorname{ch}(x-t) \varphi(t) d t=x$.
293. $\int_{0}^{x} \cos (x-t) \varphi(t) d t=x+x^{2}$.
294. $\int_{0}^{x}\left(x^{2}-t^{2}\right) \varphi(t) d t=\frac{x^{3}}{3}$.
295. $\int_{0}^{x}\left(x^{2}-4 x t+3 t^{2}\right) \varphi(t) d t=\frac{x^{4}}{12}$.
296. $\frac{1}{2} \int_{0}^{x}\left(x^{2}-4 x t+3 t^{2}\right) \varphi(t) d t=x^{2} J_{4}(2 \sqrt{x})$.
297. $\int_{0}^{x}(x-2 t) \varphi(t) d t=-\frac{x^{3}}{6}$.

Remark. If $K(x, x)=K(0) \neq 0$, then equation (1) has a solution. In problem 290, the kernel $K(x, t)$ is identically zero when $t=x$, but the solution to this equation still exists.

As already noted above, the necessary condition for the existence of a continuous solution to an integral equation of the form

$$
\int_{0}^{x} \frac{(x-t)^{n-1}}{(n-1)!} \varphi(t) d t=f(x)
$$

is that the function $f(x)$ has continuous derivatives up to the $n$-th order inclusive, and all its $n-1$ first derivatives are zero at $x=0$.

This "model" equation (4) indicates the need to match the orders of vanishing of the kernel at $t=x$ and the right-hand side $f(x)$ at $x=0$ (the right-hand side should exceed the kernel by at least 1).

Consider the integral equation

$$
\int_{0}^{x}(x-t) \varphi(t) d t=x
$$

Here $f(x)=x, n=2$. Clearly, $f(x)$ has derivatives of all orders, but its first derivative $f^{\prime}(x)=1 \neq 0$, i.e., the necessary condition is not satisfied.

Formally applying the Laplace transform to both sides of equation (5), we get

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

from which

$$
\Phi(p)=1
$$

This is the image of the $\delta$-function $\delta(x)$.

Recall that

$$
\delta(x) \risingdotseq 1, \quad \delta^{(m)}(x) \risingdotseq p^{m}
$$

$m-$ integer $\geqslant 0$.

Thus, the solution to the integral equation (5) is the $\delta$-function:

$$
\varphi(x)=\delta(x)
$$

This can be verified by direct inspection, if we consider that the convolution of the $\delta$-function with any smooth function $g(x)$ is defined as follows:

$$
\begin{aligned}
g(x) * \delta(x) & =g(x) \\
\delta^{(k)}(x) * g(x) & =g^{(k)}(x) \quad(k=1,2, \ldots)
\end{aligned}
$$

Indeed, in our case $g(x)=K(x)=x$ and

$$
\int_{0}^{x} K(x-t) \delta(t) d t=K(x)=x
$$

Thus, the solution to equation (5) exists, but in the class of generalized functions.

## Problems for Independent Solution

Solve the integral equations:

298. $\int_{0}^{x}(x-t) \varphi(t) d t=x^{2}+x-1$. 299. $\int_{0}^{x}(x-t) \varphi(t) d t=\sin x$.
299. $\int_{0}^{x}(x-t)^{2} \varphi(t) d t=x^{2}+x^{3}$.
300. $\int_{0}^{x} \sin (x-t) \varphi(t) d t=x+1$.
301. $\int_{0}^{x} \sin (x-t) \varphi(t) d t=1-\cos x$.

Integral equations of the first kind with logarithmic kernel

$$
\int_{0}^{x} \varphi(t) \ln (x-t) d t=f(x), \quad f(0)=0
$$

can also be solved using the Laplace transform.

It is known that

$$
x^{\nu} \risingdotseq \frac