Example. Restore the original $f(t)$ from its image

$$
F(p)=\frac{e^{-2 p}}{\left(p^{2}+1\right)^{2}}
$$
Solution. 1. We restore the original $r(t)$ from its image

$$
R(p)=\frac{1}{\left(p^{2}+1\right)^{2}}
$$

We have

$$
\sin t \longleftrightarrow \frac{1}{p^{2}+1}
$$

By the convolution theorem

$$
\sin t * \sin t \longleftrightarrow \frac{1}{p^{2}+1} \cdot \frac{1}{p^{2}+1}
$$

We compute the convolution

$$
\sin t * \sin t=\int_{0}^{\tau} \sin (t-\tau) \sin \tau d \tau=\frac{1}{2} t \cos t-\frac{1}{2} \sin t
$$

Thus,

$$
r(t)=\frac{1}{2} t \cos t-\frac{1}{2} \sin t
$$

2. By the delay theorem, the desired original is given by the formula

$$
f(t)=r(t-2) \eta(t-2)=\frac{1}{2}[(t-2) \cos (t-2)-\sin (t-2)] \eta(t-2)
$$

$$
\text { Answer. } f(t)=\frac{1}{2}[(t-2) \cos (t-2)-\sin (t-2)] \eta(t-2) \text {. }
$$

Conditions of the problems. Restore the original $f(t)$ from its image $F(p)$.

1. $F(p)=\frac{e^{-p}}{p^{2}+1}$.
2. $F(p)=\frac{p e^{-4 p}}{p^{2}+1}$.
3. $F(p)=\frac{e^{-7 p}}{p-3}$.
4. $F(p)=\frac{2 e^{-5 p}}{p^{2}-4}$.
5. $F(p)=\frac{e^{-3 p}}{p\left(p^{2}+9\right)}$.
6. $F(p)=\frac{e^{-\frac{p}{2}}}{p^{4}-1}$.
7. $F(p)=\frac{p e^{-p}}{p^{2}-3}$.
8. $F(p)=\frac{e^{-3 p}}{p^{2}\left(p^{2}-1\right)}$.
9. $F(p)=\frac{p e^{-6 p}}{\left(p^{2}+4\right)\left(p^{2}+1\right)}$.
10. $F(p)=\frac{p e^{-\pi p}}{\left(p^{2}+16\right)^{2}}$.

Answers. 1. $f(t)=\sin (t-1) \eta(t-1) . \quad 2$. $f(t)=\cos (t-4) \eta(t-4)$. 3. $f(t)=e^{3(t-7)} \eta(t-7)$. 4. $f(t) \operatorname{sh} 2(t-5) \eta(t-5)$. 5. $f(t)=(1-$ $-\cos (3 t-9)) \eta(t-3) / 9$. 6. $f(t)=(\operatorname{sh}(t-1 / 2)-\sin (t-1 / 2)) \eta(t-1 / 2) / 2$. 7. $f(t)=\operatorname{ch}(\sqrt{3}(t-1)) \eta(t-1)$. 8. $f(t)=(\operatorname{sh}(t-3)-t-3) \eta(t-3)$. 9. $f(t)=(\cos (t-6)-\cos (2 t-12)) \eta(t-6) .10 \cdot f(t)=(t-\pi) \sin (4 t-4 \pi) \times$ $\times \eta(t-\pi) / 8$.

### 2.12. Solution of linear

## differential equations

Problem Statement. Find the solution to the Cauchy problem for an ordinary linear differential equation with constant coefficients

$$
a_{n} x^{(n)}+a_{n-1} x^{(n-1)}+\ldots+a_{1} x^{\prime}+a_{0} x=f(t)
$$

with initial conditions

$$
x(0)=x_{0}, \quad x^{\prime}(0)=x_{0}^{\prime}, \quad \ldots, \quad x^{(n-1)}(0)=x_{0}^{(n-1)}
$$

Plan of Solution. If $f(t)$ is the original, then the desired solution $x(t)$ of the differential equation is also an original. Denote its image by $X(p)$.

1. Find the image of the left-hand side of the equation.

By the theorem on differentiation of the original

$x^{\prime}(t) \longleftrightarrow p X(p)-x(0), \quad x^{\prime \prime}(t) \longleftrightarrow p^{2} X(p)-p x(0)-x^{\prime}(0), \quad \ldots$,

$x^{(n)}(t) \longleftrightarrow p^{n} X(p)-p^{n-1} x(0)-\ldots-x^{(n-1)}(0)$.

By the linearity property

$a_{n} x^{(n)}+a_{n-1} x^{(n-1)}+\ldots+a_{1} x^{\prime}+a_{0} x \longleftrightarrow$

$\longleftrightarrow\left(a_{n} p^{n}+a_{n-1} p^{n-1}+\ldots+a_{1} p+a_{0}\right) X(p)-p^{n-1} x_{0}-p^{n-2} x_{0}^{\prime}-\ldots-x_{0}^{(n-1)}$.

2. Find the image of the right-hand side of the equation

$$
f(t) \longleftrightarrow F(p)
$$

3. Formulate the operator equation

$$
\begin{aligned}
& \left(a_{n} p^{n}+a_{n-1} p^{n-1}+\ldots+a_{1} p+a_{0}\right) X(p)= \\
& \quad=F(p)+p^{n-1} x_{0}+p^{n-2} x_{0}^{\prime}+\ldots+x_{0}^{(n-1)}
\end{aligned}
$$

4. Solve the operator equation for $X(p)$.
5. Restore the original $x(t)$ from the found image $X(p)$.
6. Check if $x(t)$ satisfies the original differential equation and initial conditions.

Write the answer.