Example 7. Reduce the equation

$$
x^{2} \frac{d^{2} y}{d x^{2}}-3 x \frac{d y}{d x}+\left(x^{4}-12\right) y=0
$$

to a Bessel equation and find its general solution.
Solution. In our case, the coefficients are $a=-3, b=-12, c=1$, $m=\overline{4, \text { therefore }}$

$$
\begin{aligned}
\alpha=\frac{a-1}{2} & =-2, \quad \beta=\frac{m}{2}=2, \quad-\frac{\alpha}{\beta}=1, \quad \frac{1}{\beta}=\frac{1}{2} \\
\gamma & =\frac{2 \sqrt{c}}{m}=\frac{1}{2}, \quad p^{2}=\frac{(a-1)^{2}-4 b}{m^{2}}=4
\end{aligned}
$$

We introduce a new independent variable $t$ and the function $u$ by the formulas

$$
\begin{aligned}
& y=\left(\frac{t}{\gamma}\right)^{-\alpha / \beta} u, \text { or } y=2 u t, \quad \text { where } u=u(t) \\
& x=\left(\frac{t}{\gamma}\right)^{-1 / \beta}, \text { or } x=\sqrt{2 t}
\end{aligned}
$$

then

$$
\frac{d y}{d x}=\frac{d(2 u t) / d t}{d x / d t}=\sqrt{2 t} \frac{d}{d t}(2 u t)=2 \sqrt{2}\left(t^{3 / 2} \frac{d u}{d t}+t^{1 / 2} u\right)
$$

Similarly, we find

$$
\frac{d^{2} y}{d x^{2}}=4 t^{2} \frac{d^{2} u}{d t^{2}}+10 t \frac{d u}{d t}+2 u
$$

Substituting into (44) instead of $x, y, \frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}$ their expressions in terms of $t$ and $u$, we obtain the Bessel equation

$$
t^{2} \frac{d^{2} u}{d t^{2}}+t \frac{d u}{d t}+\left(t^{2}-4\right) u=0
$$

the general solution of which is

$$
u=C_{1} J_{2}(t)+C_{2} Y_{2}(t)
$$

Transitioning to the variables $x$ and $y$ using the formulas $t=\frac{x^{2}}{2}, u=\frac{y}{x^{2}}$, which are obtained from (45) and (46), we get the general solution of the given equation

$$
y=x^{2}\left[C_{1} J_{2}\left(\frac{x^{2}}{2}\right)+C_{2} Y_{2}\left(\frac{x^{2}}{2}\right)\right]
$$

## Problems for Independent Solution

Find the general solutions of the following Bessel equations:

744. $x^{2} y^{\prime \prime}+x y^{\prime}+\left(4 x^{2}-\frac{1}{9}\right) y=0$.
745. $x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=0$.
746. $y^{\prime \prime}+\frac{1}{x} y^{\prime}+\frac{1}{9} y=0$.
747. $y^{\prime \prime}+\frac{1}{x} y^{\prime}+4 y=0$.
748. $x^{2} y^{\prime \prime}-2 x y^{\prime}+4\left(x^{4}-1\right) y=0$.
749. $x y^{\prime \prime}+\frac{1}{2} y^{\prime}+\frac{1}{4} y=0$.
750. $y^{\prime \prime}+\frac{5}{x} y^{\prime}+y=0$.
751. $y^{\prime \prime}+\frac{3}{x} y^{\prime}+4 y=0$.
$3^{\circ}$. Finding Periodic Solutions of Linear Differential Equations. Let a linear inhomogeneous differential equation of the second order with constant coefficients be given

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

where $f(x)$ is a periodic function with period $2 \pi$, expandable into a Fourier series

$$
f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n x+b_{n} \sin n x\right)
$$

We seek the periodic solution of equation (47) in the form

$$
y(x)=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left(A_{n} \cos n x+B_{n} \sin n x\right)
$$

Substituting the series (49) into equation (47) and choosing its coefficients so that equality (47) is formally satisfied. By equating the free terms and the coefficients of $\cos n x$ and $\sin n x$ in the left and right parts of the obtained equality, we find

$$
\begin{aligned}
A_{0}=\frac{a_{0}}{p_{2}}, \quad A_{n} & =\frac{\left(p_{2}-n^{2}\right) a_{n}-p_{1} n b_{n}}{\left(p_{2}-n^{2}\right)^{2}+p_{1}^{2} n^{2}}, \quad n=1,2, \ldots \\
B_{n} & =\frac{\left(p_{2}-n^{2}\right) b_{n}-p_{1} n a_{n}}{\left(p_{2}-n^{2}\right)^{2}+p_{1}^{2} n^{2}}
\end{aligned}
$$

The first of the equalities (50) gives the necessary condition for the existence of a solution of the form (49): if $a_{0} \neq 0$, then it is necessary that $p_{2} \neq 0$. Substituting (50) into (49), we get

$$
y(x)=\frac{a_{0}}{2 p_{2}}+\sum_{n=1}^{\infty} \frac{\left[\left(p_{2}-n^{2}\right) a_{n}-p_{1} n b_{n}\right] \cos n x+\left[\left(p_{2}-n^{2}\right) b_{n}-p_{1} n a_{n}\right] \sin n x}{\left(p_{2}-n^{2}\right)^{2}+p_{1}^{2} n^{2}}
$$

When $p_{1}=0$ and $p_{2}=n^{2}$, where $n=1,2, \ldots$, a periodic solution will exist only under the condition

$$
a_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cos n x d x=0, \quad b_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \sin n x d x=0
$$

The coefficients $A_{k}$ and $B_{k}$ for $k \neq n$ diverge according to formulas (50), while the coefficients $A_{n}$ and $B_{n}$ remain arbitrary, since the expression $A_{n} \cos n x+B_{n} \sin n x$ is the general solution of the corresponding homogeneous equation.

In the case of non-fulfillment of conditions (52), equation (47) has no periodic solutions (resonance occurs). When $p_{2}=0$ and $a_{0}=0$, the coefficient $A_{0}$ remains undefined, and equation (47) has an infinite set of periodic solutions differing from each other by a constant term.

If the right-hand side $f(x)$ of equation (47) has a period $2 l \neq 2 \pi$, then $f(x)$ must be expanded over the period $2 l$ and the solution of equation (47) must be sought in the form

$$
y=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left(A_{n} \cos \frac{n \pi x}{l}+B_{n} \sin \frac{n \pi x}{l}\right)
$$

The formulas (50) will accordingly change.