Example 4. In problems concerning the oscillations of a thin elastic plate, we arrive at the following integral equation:

$$
\psi(t)=\frac{1}{2 b t} \int_{0}^{+\infty} x f(x) \sin \frac{x^{2}}{4 b t} d x
$$

where $f(x)$ is the unknown function, and $\psi(t)$ is the known function.
Solution. Equation (8) is an integral equation of the first kind. Setting

$$
t=\frac{1}{4 b a}, \quad x^{2}=v
$$

we transform equation (8) into the form

$$
\frac{1}{\alpha} \psi\left(\frac{1}{4 b a}\right)=\int_{0}^{+\infty} f(\sqrt{v}) \sin \alpha v d v
$$

Using the inversion formula for the sine Fourier transform, we obtain

$$
f(\sqrt{v})=\frac{2}{\pi} \int_{0}^{+\infty} \psi\left(\frac{1}{4 b a}\right) \frac{\sin \alpha v}{\alpha} d \alpha
$$

or

$$
f(x)=\frac{2}{\pi} \int_{0}^{+\infty} \frac{\psi(t)}{t} \sin \frac{x^{2}}{4 b t} d t
$$

## Problems for Independent Solution

Solve the following integral equations:

230. $\int_{0}^{+\infty} \varphi(t) \cos x t d t=\frac{1}{1+x^{2}} \quad(x>0)$.
231. $\int_{0}^{+\infty} \varphi(t) \sin x t d t=f(x)$, where $f(x)= \begin{cases}\frac{\pi}{2} \sin x, & 0 \leqslant x \leqslant \pi, \\ 0, & x>\pi,\end{cases}$
232. $\int_{0}^{+\infty} \varphi(t) \cos x t d t=f(x)$, where $f(x)= \begin{cases}\cos x, & 0 \leqslant x \leqslant \pi, \\ 0, & x>\pi .\end{cases}$
233. $\int_{0}^{+\infty} \varphi(t) \cos x t d t=e^{-x} \cos x \quad(x>0)$.

## § 18. Application of the Laplace Transform to Solving Some Integral Equations

$1^{\circ}$. Integral equations of the Volterra type of convolution. Consider the Volterra integral equation of the second kind

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

where the kernel depends only on the difference $x-t$. We will call equation (1) an integral equation of the convolution type.

Let $f(x)$ and $K(x)$ be sufficiently smooth functions that grow no faster than an exponential function as $x \rightarrow \infty$, so that

$$
|f(x)| \leqslant M_{1} e^{s_{1} x}, \quad|K(x)| \leqslant M_{2} e^{s_{2} x}
$$

It can be shown that in this case the function $\varphi(x)$ will satisfy an estimate of the type (2):

$$
|\varphi(x)| \leqslant M_{3} e^{e_{3} x} .
$$

Therefore, the Laplace transform of the functions $f(x), K(x)$, and $\varphi(x)$ can be found (it will be defined in the half-plane $\operatorname{Re} p=s> \left.\max \left\{s_{1}, s_{2}, s_{3}\right\}\right)$.

Let

$$
f(x) \risingdotseq F(p), \quad \varphi(x) \risingdotseq \Phi(p), \quad K(x) \risingdotseq \bar{K}(p)
$$

Applying the Laplace transform to both sides of equation (1) and using the multiplication theorem, we find

$$
\Phi(p)=F(p)+\widetilde{K}(p) \Phi(p)
$$

From this,

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

The original $\varphi(x)$ for $\Phi(p)$ will be the solution to the integral equation (1).