Example 3. Solve the equation

$$
\begin{array}{r}
x^{\prime}+x=1 \\
x(0)=1
\end{array}
$$
Solution. Let $x(t) \risingdotseq X(p)$, then by the rule of differentiation of the original we have

$$
x^{\prime}(t) \risingdotseq p X(p)-x(0)=p X(p)-1 .
$$

It is known that $1 \risingdotseq \frac{1}{p}$, therefore, transitioning from the given problem (21)-(22) to the operator equation, we will have

$$
p X(p)-1+X(p)=\frac{1}{p}
$$

from which

$$
(p+1) X(p)=1+\frac{1}{p} \quad \text { or } \quad X(p)=\frac{1}{p}
$$

hence, $x(t) \equiv 1$.

It is easy to see that the function $x(t) \equiv 1$ satisfies the given equation and the initial condition of the problem.