Example. Find the general solution of the linear differential equation

$$
y^{\prime \prime \prime}-100 y^{\prime}=20 e^{10 x}+100 \cos 10 x
$$
Solution.

1. We write the corresponding homogeneous equation

$$
y^{\prime \prime \prime}-100 y^{\prime}=0
$$

and look for its solution in the form $y=e^{\lambda x}$, where $\lambda$ is an unknown number.

Substituting $y=e^{\lambda x}, y^{\prime}=\lambda e^{\lambda x}$, and $y^{\prime \prime}=\lambda^{2} e^{\lambda x}$ into equation (4) and canceling $e^{\lambda x}$, we obtain the characteristic equation

$$
\lambda^{3}-100 \lambda=0
$$

The characteristic equation has three roots $\lambda_{1}=0, \lambda_{2}=10$, and $\lambda_{3}=-10$.

Thus, we have the fundamental system of solutions

$$
y_{1}=1, \quad y_{2}=e^{10 x}, \quad y_{3}=e^{-10 x}
$$

and the general solution of the homogeneous equation

$$
y_{0}=C_{1}+C_{2} e^{10 x}+C_{3} e^{-10 x}
$$

2. We solve the non-homogeneous equation using the principle of superposition:

a) We look for a particular solution $Y_{1}$ of the non-homogeneous equation

$$
y^{\prime \prime \prime}-100 y^{\prime}=20 e^{10 x}
$$

in the form $Y_{1}=x A e^{10 x}$, where $A$ is an undetermined coefficient (since $a+i b=10$ is a root of the characteristic equation of multiplicity $s=1$).

Differentiating $Y_{1}$ three times and substituting into equation (5), we find $A=1 / 10$. Thus,

$$
Y_{1}=\frac{x}{10} e^{10 x}
$$

b) We look for a particular solution $Y_{2}$ of the non-homogeneous equation

$$
y^{\prime \prime \prime}-100 y^{\prime}=100 \cos 10 x
$$

in the form $Y_{2}=B_{1} \cos 10 x+B_{2} \sin 10 x$, where $B_{1}$ and $B_{2}$ are undetermined coefficients (since $a \pm i b= \pm 10 i$ are not roots of the characteristic equation, the factor $x$ is absent).

Differentiating $Y_{2}$ three times and substituting into equation (6), we find $B_{1}=0$ and $B_{2}=-1 / 20$. Thus,

$$
Y_{2}=-\frac{1}{20} \sin 10 x
$$

Using the principle of superposition (3), we obtain

$$
y=y_{0}+Y_{1}+Y_{2}=C_{1}+C_{2} e^{10 x}+C_{3} e^{-10 x}+\frac{x}{10} e^{10 x}-\frac{1}{20} \sin 10 x
$$

Answer. $y=C_{1}+C_{2} e^{10 x}+C_{3} e^{-10 x}+\frac{x}{10} e^{10 x}-\frac{1}{20} \sin 10 x$.

Conditions of the problems. Find the general solutions of the differential equations.

1. $y^{\prime \prime}+y^{\prime}=5 x+2 e^{x}$. 2. $\quad y^{\prime \prime}+2 y^{\prime}+y=e^{-x} \cos x+x e^{-x}$.
2. $y^{\prime \prime}-3 y^{\prime}=x+\cos x$. 4. $\quad y^{\prime \prime}-2 y^{\prime}+10 y=\sin 3 x+e^{x}$.
3. $y^{\prime \prime}+4 y=2 \sin 2 x-3 \cos 2 x+1 . \quad 6 . \quad y^{\prime \prime}+9 y=x e^{3 x}+2 x \sin x$.
4. $y^{\prime \prime}+y=2 x \cos x \cos 2 x . \quad$ 8. $\quad y^{\prime \prime \prime}-5 y^{\prime \prime}+8 y^{\prime}-4 y=2 e^{3 x}+3 e^{2 x}$.
5. $y^{\prime \prime \prime}+y^{\prime \prime}=x^{2}+1+3 x e^{x}$. 10. $y^{\mathrm{iv}}+2 y^{\prime \prime}+y=e^{x}+\sin x+\sin 2 x$.

Answers.

1. $y=C_{1}+C_{2} e^{-x}+e^{x}+\frac{5}{2} x^{2}-5 x$.
2. $y=\left(C_{1}+C_{2} x+\frac{x^{3}}{6}-\cos x\right) e^{-x}$.
3. $y=C_{1}+C_{2} e^{3 x}-\frac{1}{10}(\cos x+3 \sin x)-\frac{x^{2}}{6}-\frac{x}{9}$.
4. $y=\left(C_{1} \cos 3 x+C_{2} \sin 3 x\right) e^{x}+\frac{1}{37}(\sin 3 x+6 \cos 3 x)+\frac{e^{x}}{9}$.
5. $y=C_{1} \cos 2 x+C_{2} \sin 2 x-\frac{x}{4}(3 \sin 2 x+2 \cos 2 x)+\frac{1}{4}$.
6. $y=C_{1} \cos 3 x+C_{2} \sin 3 x+\frac{1}{4} x \sin x-\frac{1}{16} \cos x+\frac{1}{54}(3 x-1) e^{3 x}$.
7. $y=C_{1} \cos x+C_{2} \sin x+\frac{x}{4} \cos x+\frac{x^{2}}{4} \sin x-\frac{x}{8} \cos 3 x+\frac{3}{32} \sin 3 x$.
8. $y=C_{1} e^{x}+\left(C_{2}+C_{3} x\right) e^{2 x}+e^{3 x}+\frac{3}{2} x^{2} e^{2 x}$.
9. $y=C_{1} e^{-x}+C_{2}+C_{3} x+\frac{3}{2} x^{2}-\frac{1}{3} x^{3}+\frac{1}{12} x^{4}+\left(\frac{3}{2} x-\frac{15}{4}\right) e^{x}$.
10. $y=\left(C_{1} x+C_{2}\right) \cos x+\left(C_{3} x+C_{4}\right) \sin x+\frac{1}{4} e^{x}-\frac{1}{8} x^{2} \sin x+\frac{1}{9} \sin 2 x$.

### 11.11. Lagrange's Method

Problem Statement. Find the solution of the Cauchy problem for the linear non-homogeneous equation with constant coefficients

$$
y^{\prime \prime}+p_{1} y^{\prime}+p_{2} y=f(x)
$$

with initial conditions

$$
y\left(x_{0}\right)=y_{0}, \quad y^{\prime}\left(x_{0}\right)=y_{0}^{\prime}
$$

## PLAN OF SOLUTION.

1. We write the corresponding homogeneous equation with constant coefficients

$$
y^{\prime \prime}+p_{1} y^{\prime}+p_{2} y=0
$$

We find the fundamental system of solutions $y_{1}(x)$ and $y_{2}(x)$ and the general solution of the homogeneous equation

$$
y=C_{1} y_{1}(x)+C_{2} y_{2}(x) \text {. }
$$

2. We apply Lagrange's method (method of variation of arbitrary constants).

If the fundamental system of solutions $y_{1}(x), y_{2}(x)$ of the homogeneous equation (2) is known, then the general solution of the corresponding non-homogeneous equation (1) can be found using the formula

$$
y=C_{1}(x) y_{1}(x)+C_{2}(x) y_{2}(x),
$$

where the functions $C_{1}^{\prime}(x)$ and $C_{2}^{\prime}(x)$ are determined from the system of linear algebraic equations

$$
\left\{\begin{array}{l}
C_{1}^{\prime}(x) y_{1}(x)+C_{2}^{\prime}(x) y_{2}(x)=0 \\
C_{1}^{\prime}(x) y_{1}^{\prime}(x)+C_{2}^{\prime}(x) y_{2}^{\prime}(x)=f(x)
\end{array}\right.
$$

Integrating, we find the functions $C_{1}(x)$ and $C_{2}(x)$ and write the general solution of the non-homogeneous equation.

3. Using the initial conditions ( $1^{\prime}$ ), we find the solution of the Cauchy problem.

We write the answer in the form $y=y(x)$.