## Example. Solve the integral equation

$$
\int_{0}^{x} \cos (x-t) \varphi(t) d t=x
$$
Solution. The functions $f(x)=x, K(x, t)=\cos (x-t)$ satisfy the above formulated conditions of continuity and differentiability.

Differentiating both sides of (4) with respect to $x$, we get

$$
\varphi(x) \cos 0-\int_{0}^{x} \sin (x-t) \varphi(t) d t=1
$$

or

$$
\varphi(x)=1+\int_{0}^{x} \sin (x-t) \varphi(t) d t
$$

Equation (5) is an integral equation of the second kind of convolution type.

Applying the Laplace transform, we find its solution:

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

from which

$$
\Phi(p)=\frac{p^{2}+1}{p^{3}}=\frac{1}{p}+\frac{1}{p^{3}} \doteqdot 1+\frac{x^{2}}{2}
$$

The function $\varphi(x)=1+\frac{x^{2}}{2}$ will be the solution of equation (5), and consequently, of the original equation (4), which is easy to verify by direct substitution.

## Problems for Independent Solution

Solve the following integral equations of the first kind, first reducing them to integral equations of the second kind:
27T. $\int_{0}^{x} e^{x-t} \varphi(t) d t=\sin x$.
278. $\int_{0}^{x} 3^{x-t} \varphi(t) d t=x$.
279. $\int_{0}^{x} a^{z-t} \varphi(t) d t=f(x), \quad f(0)=0$.
280. $\int_{0}^{x}\left(1-x^{2}+t^{2}\right) \varphi(t) d t=\frac{x^{2}}{2}$.
281. $\int_{0}^{x}\left(2+x^{2}-t^{2}\right) \varphi(t) d t=x^{2}$.
282. $\int_{0}^{x} \sin (x-t) \varphi(t) d t=e^{x^{2} / 2}-1$.

283. Reduce the equation

$$
\int_{a}^{x} \frac{H(x, s)}{(x-s)^{\alpha}} \varphi(s) d s=f(x)
$$

to a Volterra equation of the second kind, assuming that $H(x, s)$ and $f(x)$ are continuously differentiable, $f(a)=0$, $H(a, a) \neq 0$ and $0<\alpha<1$.

Suppose that in the time interval from $\tau$ to $\tau+d \tau$ goods are purchased for an amount $\varphi(\tau) d \tau$. This stock decreases due to sales such that the value of the remaining stock at time $t>\tau$ is $K(t-\tau) \varphi(\tau) d \tau$. Therefore, the value of the unsold portion of the goods, acquired through purchases, will be at any time $t$ equal to

$$
\int_{0}^{t} K(t-\tau) \varphi(\tau) d \tau
$$

Thus, $\varphi(t)$ must satisfy the integral equation

$$
1-K(t)=\int_{0}^{\mathrm{t}} K(t-\tau) \varphi(\tau) d \tau
$$

We have obtained a Volterra integral equation of the first kind of convolution type.

Let $f(x)$ and $K(x)$ be original functions, and let

$$
\begin{aligned}
f(x) & \risingdotseq F(p) \\
K(x) & \risingdotseq \tilde{K}(p) \\
\varphi(x) & \risingdotseq \Phi(p)
\end{aligned}
$$

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

$$
\tilde{K}(p) \cdot \Phi(p)=F(p)
$$

from which

$$
\Phi(p)=\frac{F(p)}{\widetilde{K}(p)} \quad(\tilde{K}(p) \neq 0)
$$

The original function $\varphi(x)$ for the function $\Phi(p)$, defined by equation (2), will be the solution of the integral equation (1).