Confused in regard to Thereom about ordinary point/analytic point I am having some trouble understand the implication of the theorem

$\mathbf{Theorem}:$ If $x_o$ is an ordinary point of the ODE $P(x)y''+Q(x)y'+R(x)y=0$, ( that is $Q/P$ and $R/P$ are analytic at $x_o$) then the general solution of the ODE is of the form
$$y=\sum_{n=0}^{\infty}a_n(x-x_o)^{n}=a_oy_1(x)+a_1y_2(x)$$

Now here is where I am confused, next it says,

notice $y_1(x)=1+b_2(x-x_o)^2+..$ and $y_2=(x-x_o)+c_2(x-x_o)^2+…$ where $b_2+c_2=a_2$, hence $y_1$ satisfies $y_1(x_o)=1$ and $y'_1(x_o)=0$ and $y_2(x_o)=0$ and $y'_2(x_o)=1$

I am really having trouble understand this. How did they deduce the form of the solutions, and how can they have known what the solutions satisfy?
Thanks
 A: Please correct "analytic at $x_0$" to "analytic in a neighborhood $\mathcal{N}$ of $x_0$". Analyticity isn't a point property. If $x_0$ is a regular point of the differential equation, and if $A$, $B$ are constants, then there is a unique $y$ that is analytic in $\mathcal{N}$ and is a solution of
$$
          y''+\frac{Q}{P}y'+\frac{R}{P}y = 0,\\
            y(x_0)=A,\;\; y'(x_0)=B.
$$
This existence and uniqueness result can be obtained by Picard iteration in a complex neighbhorhood of $x_0$.
A basis of solutions is $\{ y_1, y_2\}$ which satisfy the equation and the conditions
$$  
    y_1(x_0)=1,\; y_1'(x_0)=0 \\
    y_2(x_0)=0,\; y_2'(x_0)=1.
$$
To see why this is a basis, suppose $y$ is a solution of the equation. Then
$$
                      w(x)=y(x_0)y_1(x) + y'(x_0)y_2(x)
$$
is also a solution of the equation with $w(x_0)=y(x_0)$, $w'(x_0)=y'(x_0)$. By uniqueness, $w = y$.
Knowing that $y_1$ and $y_2$ have McClaurin series expansion about $x_0$, it is easy to verify that the constant coefficient in the series expansion of $y_2$ is $0$ because $y_2(x_0)=0$; and the coefficient of $(x-x_0)$ is $1$ because $y_2'(x_0)=1$. Likewise, the constant coefficient for $y_1$ is $1$ and the coefficient of $(x-x_0)$ for $y_1$ is $0$.
