Example 3. Solve the boundary value problem $x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0, y(1)=1$, $y(x)$ is bounded as $x \rightarrow 0$.
Solution. The given equation is an Euler equation. Its general solution has the form $y(x)=\frac{C_{1}}{x^{3}}+C_{2} x^{2}$ (see, for example, 1, item $4^{\circ}, \S$ 15). By the condition, the solution $y(x)$ must be bounded as $x \rightarrow 0$. This requirement will be satisfied if in the general solution we set $C_{1}=0$. Then we will have $y(x)=C_{2} x^{2}$. The boundary condition $y(1)=1$ gives $C_{2}=1$. Therefore, the desired solution is $y=x^{2}$.

## Problems for Independent Solution

706. For which values of $\lambda$ does the equation $y^{\prime \prime}+\lambda y=0$ have a non-zero solution satisfying the conditions:
a) $y^{\prime}(0)=y^{\prime}(\pi)=0$,
b) $y(0)=y(\pi), y^{\prime}(0)=y^{\prime}(\pi)$ ?
707. For which values of $\lambda$ does the boundary value problem $y^{\prime \prime}+\lambda y=0, y(0)=y(1)=0$ have a trivial solution $y \equiv 0$ ?
708. Which of the following boundary value problems is solvable:
a) $y^{\prime \prime}-y=0$
$y(0)=0, y(2 \pi)=1$
b) $y^{\prime \prime}+y=0$
$y(0)=0, y(2 \pi)=1$ ?
709. Solve the boundary value problem $y^{\prime \prime}+\left(\lambda-\omega^{2}\right) y=0, y(0)=y(1), y^{\prime}(0)=y^{\prime}(1)$. Consider the cases $\lambda-\omega^{2}>0, \lambda-\omega^{2}=0, \lambda-\omega^{2}<0$.
710. Find the solution of the equation $y y^{\prime \prime}+\left(y^{\prime}\right)^{2}+1=0$ passing through the points $(0,1)$ and $(1,2)$.

Solve the following boundary value problems:

711. $y^{\prime \prime}+y=0$, $y(0)=0, y\left(\frac{\pi}{2}\right)=\alpha$.
712. $y^{\prime \prime}-y=0$, $y(0)=0, y^{\prime}(1)=1$.
713. $y^{\prime \prime}-2 y^{\prime}+2 y=0$, $y(0)=0, \quad y^{\prime}(\pi)=e^{\pi}$
714. $y^{\prime \prime}+\alpha y^{\prime}=0$, $y(0)=e^{\alpha}, \quad y^{\prime}(1)=0$.
715. $y^{\prime \prime}+\alpha^{2} y=1$, $y^{\prime}(0)=\alpha, \quad y^{\prime}(\pi)=0 \quad(0<\alpha<1)$.
716. $y^{\prime \prime}+y=1$,

$y(0)=0, \quad y^{\prime}(\pi)=0$.

717. $y^{\prime \prime}+\lambda^{2} y=0$, $y^{\prime}(0)=0, \quad y^{\prime}(\pi)=0$.
718. $y^{\prime \prime}+\lambda^{2} y=0$,

$y(0)=0, \quad y^{\prime}(\pi)=0$.

719. $y^{\prime \prime \prime}+y^{\prime \prime}-y^{\prime}-y=0$,

$y(0)=-1, \quad y^{\prime}(0)=2, \quad y(1)=0$.

720. $y^{\text {IV }}-\lambda^{4} y=0$,

$y(0)=y^{\prime \prime}(0)=0, \quad y(\pi)=y^{\prime \prime}(\pi)=0$,

721. $x y^{\prime \prime}+y^{\prime}=0$, $y(1)=\alpha y^{\prime}(1), y(x)$ is bounded as $x \rightarrow 0$.
722. $x^{2} y^{\text {IV }}+4 x y^{\prime \prime \prime}+2 y^{\prime \prime}=0$

$y(1)=y^{\prime}(1)=0, \quad y(x)$ is bounded as $x \rightarrow 0$.

723. $x^{3} y^{\mathrm{IV}}+6 x^{2} y^{\prime \prime \prime}+6 x y^{\prime \prime}=0$,

$y(1)=y^{\prime}(1)=0, y(x)$ is bounded as $x \rightarrow 0$.

## § 18. Integration of Differential Equations Using Series

$1^{\circ}$. Expansion of the solution in a power series. This method is particularly convenient for linear differential equations. Let us illustrate its application using a second-order differential equation. Suppose we have a second-order differential equation

$$
y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0
$$

Assume that the coefficients $p(x)$ and $q(x)$ can be represented as series in positive integer powers of $x$, so that equation (1) can be rewritten as

$$
y^{\prime \prime}+\left(a_{0}+a_{1} x+a_{2} x^{2}+\ldots\right) y^{\prime}+\left(b_{0}+b_{1} x+b_{2} x^{2}+\ldots\right) y=0
$$

We will seek the solution of this equation in the form of a power series

$$
y=\sum_{k=0}^{\infty} c_{k} x^{k}
$$

Substituting this expression for $y$ and its derivatives into (2), we get

$$
\sum_{k=2}^{\infty} k(k-1) c_{k} x^{k-2}+\sum_{k=0}^{\infty} a_{k} x^{k} \sum_{k=1}^{\infty} k c_{k} x^{k-1}+\sum_{k=0}^{\infty} b_{k} x^{k} \sum_{k=0}^{\infty} c_{k} x^{k}=0
$$

Multiplying the power series, collecting like terms, and equating the coefficients of all powers of $x$ in the left-hand side (4) to zero, we obtain a series of equations:

$$
\begin{array}{c|l}
x^{0} & 2 \cdot 1 c_{2}+a_{0} c_{1}+b_{0} c_{0}=0 \\
x^{1} & 3 \cdot 2 c_{3}+2 a_{0} c_{2}+a_{1} c_{1}+b_{0} c_{1}+b_{1} c_{0}=0 \\
x^{2} & 4 \cdot 3 c_{4}+3 a_{0} c_{3}+2 a_{1} c_{2}+a_{2} c_{1}+b_{0} c_{2}+b_{1} c_{1}+b_{2} c_{0}=0 \\
\ldots & \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\end{array}
$$

Each subsequent equation (5) contains one more unknown coefficient than the previous one. The coefficients $c_{0}$ and $c_{1}$ remain arbitrary and play the role of arbitrary constants. The first equation (5) gives $c_{2}$, the second gives $c_{3}$, the third gives $c_{4}$, and so on. In general, from the $(k+1)$-th equation, we can determine $c_{k+2}$, knowing $c_{0}, c_{1}, \ldots, c_{k+1}$.

Practically, it is convenient to proceed as follows. We will determine two solutions $y_{1}(x)$ and $y_{2}(x)$ according to the above scheme, choosing $c_{0}=1$ and $c_{1}=0$ for $y_{1}(x)$, and $c_{0}=0$ and $c_{1}=1$ for $y_{2}(x)$, which is equivalent to the following initial conditions:

$$
y_{1}(0)=1, \quad y_{1}^{\prime}(0)=0, \quad y_{2}(0)=0, \quad y_{2}^{\prime}(0)=1
$$

Any solution of equation (1) will be a linear combination of the solutions $y_{1}(x)$ and $y_{2}(x)$.

If the initial conditions are of the form $y(0)=A, y^{\prime}(0)=B$, then obviously,

$$
y=A y_{1}(x)+B y_{2}(x)
$$

The following theorem holds.

Theorem. If the series

$$
p(x)=\sum_{k=0}^{\infty} a_{k} x^{k} \quad \text { and } \quad q(x)=\sum_{k=0}^{\infty} b_{k} x^{k}
$$

converge for $|x|<R$, then the power series (3) constructed by the above method will also converge for these values of $x$ and will be a solution of equation (1).

In particular, if $p(x)$ and $q(x)$ are polynomials in $x$, then the series (3) will converge for any value of $x$.