Example 8. Find the general solution of the equation

$$
y^{\prime \prime}-12 y^{\prime}+36 y=\sin 3 x
$$
Solution. The right-hand side of the given equation is a trigonometric polynomial of the form $f(x)=a \cos m x+b \sin m x$, where $a=0, b=1$, $m=3$. Since condition (10.28) is satisfied, we seek a particular solution in the form

$$
\bar{y}=A \cos 3 x+B \sin 3 x
$$

We find the derivatives

$$
\bar{y}^{\prime}=-3 A \sin 3 x+3 B \cos 3 x, \bar{y}^{\prime \prime}=-9 A \cos 3 x-9 B \sin 3 x
$$

Substituting the expressions for $\bar{y}, \bar{y}^{\prime}, \bar{y}^{\prime \prime}$ into the given equation, we get

$$
\begin{gathered}
(-9 A \cos 3 x-9 B \sin 3 x)-12(-3 A \sin 3 x+3 B \cos 3 x)+ \\
+36(A \cos 3 x+B \sin 3 x) \equiv \sin 3 x
\end{gathered}
$$

or

$$
(27 A-36 B) \cos 3 x+(27 B+36 A) \sin 3 x \equiv \sin 3 x
$$

from which

$$
27 A-36 B=0,27 B+36 A=1
$$

Solving the obtained system of equations, we find

$$
A=\frac{4}{225}, B=\frac{1}{75}
$$

Thus, we have the particular solution

$$
\bar{y}=\frac{4}{225} \cos 3 x+\frac{1}{75} \sin 3 x
$$

The corresponding homogeneous equation

$$
y^{\prime \prime}-12 y^{\prime}+36 y=0
$$

has the general solution

$$
y_{0}=\left(C_{1}+C_{2} x\right) e^{6 x}
$$

since for the characteristic equation

$$
k^{2}-12 k+36=0
$$

we get $k_{1}=k_{2}=6$.

By formula (10.21), we find the general solution

$$
y=\left(C_{1}+C_{2} x\right) e^{6 x}+\frac{4}{225} \cos 3 x+\frac{1}{75} \sin 3 x
$$

of the original equation.

Remark. We sought the particular solution in the form $\bar{y}=A \cos 3 x+ B \sin 3 x$, and not in the form $\bar{y}=B \sin 3 x$, because the fact that $a=0$ does not imply that $A=0$.

## Problems

1. $y^{\prime \prime}-6 y^{\prime}-7 y=32 e^{3 x}$.
2. $y^{\prime \prime}+2 y^{\prime}-8 y=12 e^{2 x}$.
3. $y^{\prime \prime}-2 y^{\prime}+y=e^{x}$.
4. $y^{\prime \prime}+7 y^{\prime}+12 y=24 x^{2}+16 x-15$.
5. $y^{\prime \prime}+y^{\prime}-6 y=-x^{2}-\frac{29}{18}$.
6. $y^{\prime \prime}-4 y^{\prime}+3 y=x-1$.
7. $y^{\prime \prime}-2 y^{\prime}=6 x^{2}-10 x+12$.
8. $y^{\prime \prime}-3 y^{\prime}-10 y=\sin x+3 \cos x$.
9. $y^{\prime \prime}-8 y^{\prime}+12 y=-65 \cos 4 x$. 10. $y^{\prime \prime}+16 y=-24 \sin 4 x$.

## Answers

1. $y=C_{1} e^{7 x}+C_{2} e^{-x}-2 e^{3 x}$.
2. $y=C_{1} e^{-4 x}+C_{2} e^{2 x}+2 x e^{2 x}$.
3. $y=$
$=\left(C_{1}+C_{2} x\right) e^{x}+\frac{1}{2} x^{2} e^{x}$
4. $y=C_{1} e^{-3 x}+C_{2} e^{-4 x}+2 x^{2}-x-1$.
5. $y=$ $=C_{1} e^{2 x}+C_{2} e^{-3 x}+\frac{1}{6} x^{2}+\frac{1}{18} x+\frac{1}{3}$.
6. $y=C_{1} e^{3 x}+C_{2} e^{x}+\frac{1}{3} x+\frac{1}{9}$. 7. $y=$ $=C_{1}+C_{2} e^{2 x}-x^{3}+x^{2}-5 x . \quad$ 8. $y=C_{1} e^{5 x}+C_{2} e^{-2 x}-\frac{2}{13} \sin x-\frac{9}{39} \cos x$.
7. $y=C_{1} e^{6 x}+C_{2} e^{2 x}+2 \sin 4 x+\frac{1}{4} \cos 4 x . \quad$ 10. $y=C_{1} \sin 4 x+C_{2} \cos 4 x+$ $+3 x \cos 4 x$.

## Chapter 11.

Differential Equations of Order Higher than Second. Systems of Differential Equations

## § 11.1. Equations Admitting Order Reduction

An equation of the form

$$
y^{(n)}=f(x)
$$

is solved by $n$-fold integration.

The equation

$$
F\left(x, y^{(k)}, y^{(k+1)}, \ldots, y^{(n)}\right)=0
$$

not containing the unknown function $y$, by the substitution

$$
y^{(k)}=u
$$

where $y^{(k)}$ is the lowest of the derivatives, is reduced to the equation

$$
F\left(x, u, u^{\prime}, u^{\prime \prime}, \ldots, u^{(n-k)}\right)=0
$$

the order of which is $n-k$.

The equation

$$
F\left(y, y^{\prime}, \ldots, y^{(k)}, y^{(k+1)}, \ldots, y^{(n)}\right)=0
$$

not containing the independent variable $x$, also admits order reduction using the substitution

$$
y^{\prime}=p
$$