## Example 4. Solve the equation

$$
2 x^{2} y^{\prime \prime}+\left(3 x-2 x^{2}\right) y^{\prime}-(x+1) y=0
$$
Solution. Rewrite (27) as

$$
y^{\prime \prime}+\frac{3 x-2 x^{2}}{2 x^{2}} y^{\prime}-\frac{x+1}{2 x^{2}} y=0
$$

or

$$
y^{\prime \prime}+\frac{3-2 x}{2 x} y^{\prime}-\frac{x+1}{2 x^{2}} y=0
$$

The solution $y(x)$ will be sought in the form

$$
y(x)=x^{\rho} \sum_{k=0}^{x} C_{k} x^{k} \quad\left(C_{0} \neq 0\right)
$$

To find $\rho$, we write the defining equation

$$
\rho(\rho-1)+a_{0} \rho+b_{0}=0
$$

where

$$
a_{0}=\lim _{x \rightarrow 0} \frac{3-2 x}{2}=\frac{3}{2}, \quad b_{0}=\lim _{x \rightarrow 0}\left(-\frac{x+1}{2}\right)=-\frac{1}{2}
$$

Thus,

$$
\rho(\rho-1)+\frac{3}{2} \rho-\frac{1}{2}=0, \quad \text { or } \quad \rho^{2}+\frac{1}{2} \rho-\frac{1}{2}=0
$$

from which $\rho_{1}=\frac{1}{2}, \quad \rho_{2}=-1$.

According to the given rule, we take

$$
y_{1}(x)=x^{1 / 2} \sum_{k=0}^{\infty} C_{k} x^{k} \quad(x>0) ; \quad y_{2}(x)=\frac{1}{x} \sum_{k=0}^{\infty} A_{k} x^{k}
$$

To find $C_{0}, C_{1}, \ldots, C_{n}, \ldots$, we need to substitute $y_{1}(x)$ and its derivatives $y_{1}^{\prime}(x)$ and $y_{1}^{\prime \prime}(x)$ into equation $(27)$:

$$
\begin{aligned}
y_{1}(x)= & \sum_{k=0}^{x} C_{k} x^{k+(1 / 2)}, \quad y_{1}^{\prime}(x)=\sum_{k=0}^{x}\left(k+\frac{1}{2}\right) C_{k} x^{k-(1 / 2)} \\
& y_{1}^{\prime \prime}(x)=\sum_{k=0}^{x}\left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right) C_{k} x^{k-(3 / 2)}
\end{aligned}
$$

Substituting $y_{1}(x), y_{1}^{\prime}(x), y_{1}^{\prime \prime}(x)$ into (27) gives

$$
\begin{aligned}
& 2 x^{2} \sum_{k=0}^{\infty}\left(k^{2}-\frac{1}{4}\right) C_{k} x^{k-(3 / 2)}+\left(3 x-2 x^{2}\right) \sum_{k=0}^{\infty}\left(k+\frac{1}{2}\right) C_{k} x^{k-(1 / 2)}- \\
&-(x+1) \sum_{k=0}^{\infty} C_{k} x^{k+(1 / 2)}=0
\end{aligned}
$$

After transformations, (28) can be rewritten as:

$$
\sum_{k=0}^{\infty} k(2 k+3) C_{k} x^{k+(1 / 2)}-\sum_{k=0}^{\infty} 2(k+1) C_{k} x^{k+(3 / 2)}=0
$$

Since the solution is sought for $x>0$, (29) can be simplified by dividing by $x^{1 / 2}$, which gives

$$
\sum_{k=1}^{\infty} k(2 k+3) C_{k} x^{k}-\sum_{k=0}^{\infty} 2(k+1) C_{k} x^{k+1}=0
$$

From this, we find the relation for determining the coefficients

| $x^{\prime}$ | $1 \cdot 5 C_{1}-2 \cdot 1 C_{0}=0$, |
| :--- | :--- |
| $x^{2}$ | $2 \cdot 7 C_{2}-2 \cdot 2 C_{1}=0$, |
| $x^{3}$ | $3 \cdot 9 C_{3}-2 \cdot 3 C_{2}=0$, |
| $\ldots$ | $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ |
| $x^{n}$ | $n(2 n+3) C_{n}-2 n C_{n-1}=0$, |
| $\ldots$ | $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ |

Assuming $C_{0}=1$ in the first equation of the relations (31), we get $C_{1}=\frac{2}{5}$. From the second equation, we have $C_{2}=\frac{2^{2}}{5 \cdot 7}$. From the third, $C_{3}=\frac{2^{3}}{5 \cdot 7 \cdot 9}$, and so on. It is easy to see that

$$
C_{n}=\frac{2^{n}}{5 \cdot 7 \cdot 9 \ldots(2 n+3)} \quad(n=1,2,3, \ldots)
$$

Thus,

$$
y_{1}(x)=x^{1 / 2}\left[1+\sum_{k=1}^{x} \frac{(2 x)^{k}}{5 \cdot 7 \cdot 9 \ldots(2 k+3)}\right]
$$

Similarly, we find the coefficients $\boldsymbol{A}_{k}$. It turns out that for $\boldsymbol{A}_{0}=1$

$$
A_{1}=1, \quad A_{2}=\frac{1}{2!}, \quad \ldots, \quad A_{k}=\frac{1}{k!},
$$

so that

$$
y_{2}(x)=\frac{1}{x} \sum_{k=0}^{x} \frac{x^{k}}{k!}, \text { or } y_{2}(x)=\frac{e^{x}}{x} .
$$

The general solution of equation (27) is $y(x)=A y_{1}(x)+B y_{2}(x)$, where $A$ and $B$ are arbitrary constants, and the functions $y_{1}(x)$ and $y_{2}(x)$ are given by formulas (32) and (33).