Example 5. Solve the equation $y^{\prime \prime \prime}+y^{\prime}=\operatorname{tg} x$.
Solution. The characteristic equation $r^{3}+r=0$ has roots $r_{1}=0, r_{2}=i, r_{3}=-i$, therefore

$$
y_{1}=1, y_{2}=\cos x, y_{3}=\sin x
$$

and the general solution of the homogeneous equation is expressed by the formula

$$
y_{0}=C_{1}+C_{2} \cos x+C_{3} \sin x
$$

Using the method of variation of arbitrary constants, we find the general solution of the given non-homogeneous equation, assuming

$$
y=C_{1}(x)+C_{2}(x) \cos x+C_{3}(x) \sin x
$$

The system (11.27) for determining the functions $C_{1}(x), C_{2}(x), C_{3}(x)$ is written as follows:

$$
\begin{gathered}
C_{1}^{\prime}+C_{2}^{\prime} \cos x+C_{3}^{\prime} \sin x=0 \\
-C_{2}^{\prime} \sin x+C_{3}^{\prime} \cos x=0 \\
-C_{2}^{\prime} \cos x-C_{3}^{\prime} \sin x=\operatorname{tg} x
\end{gathered}
$$

Multiplying both sides of the second equation by $\sin x$, the third by $\cos x$, and adding term by term, we get

$$
C_{2}^{\prime}=-\sin x
$$

From the second equation, we have

$$
C_{3}^{\prime}=-\frac{\sin ^{2} x}{\cos x}
$$

Adding term by term the first equation with the third, we get

$$
C_{1}^{\prime}=\operatorname{tg} x
$$

Integrating the three obtained equations, we find:

$$
\begin{gathered}
C_{1}(x)=-\ln \cos x+C_{1} \\
C_{2}(x)=\cos x+C_{2} \\
C_{3}(x)=\sin x-\ln \operatorname{tg}\left(\frac{x}{2}+\frac{\pi}{4}\right)+C_{3}
\end{gathered}
$$

where $C_{1}, C_{2}, C_{3}$ are arbitrary constants.

Therefore, the general solution of the given non-homogeneous equation is expressed by the formula

$$
\begin{aligned}
y= & -\ln \cos x-\sin x \cdot \ln \operatorname{tg}\left(\frac{x}{2}+\frac{\pi}{4}\right)+ \\
& +C_{2} \cos x+C_{3} \sin x+\bar{C}_{1}
\end{aligned}
$$

where $\bar{C}_{1}=C_{1}+1$.

## Problems

Solve the linear non-homogeneous equations with constant coefficients:

1. $y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=(x+1) e^{2 x}$.
2. $y^{\prime \prime \prime}-7 y^{\prime}+6 y=x^{2}$.
3. $y^{\mathrm{IV}}-8 y^{\prime \prime \prime}+23 y^{\prime \prime}-28 y^{\prime}+12 y=x$.
4. $y^{\mathrm{IV}}-2 y^{\prime \prime}+y=-8 e^{-x}+8 e^{x}+12 \sin x-12 \cos x$.
5. $y^{(5)}-y^{\prime \prime}=x$.
6. $y^{(6)}+3 y^{\mathrm{IV}}+3 y^{\prime \prime}+y=9 \sin 2 x$.
7. $y^{\prime \prime \prime}+2 y^{\prime \prime}+5 y^{\prime}=4 x e^{-x}-68 \cos 2 x+x$.
8. $y^{\mathrm{IV}}-3 y^{\prime \prime}-4 y=x^{2}+1+e^{3 x}+4 \cos x$.

Solve the linear non-homogeneous equations using the method of variation of arbitrary constants:
9. $y^{\prime \prime \prime}-3 y^{\prime \prime}+y^{\prime}-3 y=\sin x$.
10. $y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=\frac{2 x^{3}+x^{2}-4 x-6}{x^{4}}$.

## Answers

1. $y=C_{1} e^{x}+C_{2} \cos x+C_{3} \sin x+e^{2 x}\left(\frac{x}{5}-\frac{4}{25}\right)$ 2. $y=\frac{1}{6}\left(x^{2}+\frac{7}{3} x+\right.$ $\left.+\frac{49}{18}\right)+C_{1} e^{x}+C_{2} e^{2 x}+C_{3} e^{-3 x}$. 3. $y=\frac{3 x+7}{36}+\left(C_{1}+C_{2} x\right) e^{2 x}+C_{3} e^{x}+$ $+C_{4} e^{3 x} . \quad$ 4. $y=\left(C_{1}+C_{2} x+x^{2}\right) e^{x}+\left(C_{3}+C_{4} x-x^{2}\right) e^{-x}+3 \sin x-3 \cos x$. 5. $y=C_{1}+C_{2} x+\frac{x^{3}}{6}+C_{3} e^{-x}+C_{4} e^{-\frac{x}{2}} \cos \frac{\sqrt{3}}{2} x+C_{5} e^{-\frac{x}{2}} \sin \frac{\sqrt{3}}{2} x . \quad$ 6. $y=$ $=\left(C_{1}+C_{2} x+C_{3} x^{2}\right) \cos x+\left(C_{4}+C_{5} x+C_{6} x^{2}\right) \sin x-\frac{1}{3} \sin 2 x . \quad$ 7. $y=-(x+$ +1) $e^{-x}+8 \cos 2 x-2 \sin 2 x+\frac{x^{2}}{10}-\frac{2 x}{25}+C_{1}+C_{2} e^{-x} \cos 2 x+C_{3} e^{-x} \sin 2 x$.
2. $y=C_{1} e^{2 x}+C_{2} e^{-2 x}+C_{3} \cos x+C_{4} \sin x-\frac{x^{2}}{4}+\frac{1}{8}+\frac{e^{3 x}}{50}-\frac{2 x \sin x}{5}$. 9. $y=$ $=C_{1} e^{3 x}+C_{2} \sin x+C_{3} \cos x+\frac{3 \cos x-21 \sin x}{200}+\frac{x}{20}(3 \cos x-\sin x)$. 10. $y=$ $=C_{1} e^{x}+C_{2} e^{-x}+C_{3} e^{2 x}+\frac{1}{x}$.

## § 11.4. Systems of Linear Equations with Constant Coefficients

A system of linear differential equations of the first order with constant coefficients is a set of differential equations:

$$
\left.\begin{array}{l}
y_{1}^{\prime}=a_{11} y_{1}+a_{12} y_{2}+\ldots+a_{1 n} y_{n}+f_{1}(x) \\
y_{2}^{\prime}=a_{21} y_{1}+a_{22} y_{2}+\ldots+a_{2 n} y_{n}+f_{2}(x) ; \\
\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
y_{n}^{\prime}=a_{n 1} y_{1}+a_{n 2} y_{2}+\ldots+a_{n} y_{n}+f_{n}(x)
\end{array}\right\}
$$

where $y_{1}, y_{2}, \ldots, y_{n}$ are unknown functions of the independent variable $x$; $a_{i k}(i, k=1,2, \ldots, n)$ are constant values.

If for all considered values of $x$ all given functions $f_{k}(x) \equiv 0(k=1,2, \ldots, n)$, then the system (11.28) is called homogeneous, otherwise it is called non-homogeneous.

A solution of the system (11.28) is a system of $n$ functions

$$
y_{1}=y_{1}(x), y_{2}=y_{2}(x), \ldots, y_{n}=y_{n}(x)
$$

that turn the equalities (11.28) into identities.

Cauchy Problem. Find such a solution (11.29) of the system (11.28) that at $x=x_{0}$ it takes the given values

$$
y_{1}\left(x_{0}\right)=b_{1}, y_{2}\left(x_{0}\right)=b_{2}, \ldots, y_{n}\left(x_{0}\right)=b_{n}
$$

By the method of elimination of $(n-1)$ unknown functions, the system (11.28) in some cases can be reduced to a differential equation of the $n$-th order with respect to one of the unknown functions.

Remark. If the independent variable is denoted by $t$, then the system (11.28) will take the form

$$
\dot{y}_{k}=\sum_{l=1}^{n} a_{k l} y_{l}+f_{k}(t)(k=1,2,3, \ldots, n)
$$

where the dot denotes the derivative with respect to $t$. This notation is often used in mechanics.