While using the Frobenius method to solve a second order ODE of the form
$$y^{\prime \prime} + p(x)y^\prime + q(x)y = 0$$
if the roots of the indicial equation $(r(r-1)+p_0r+q_0=0)$ $r_1, r_2 (r_1>r_2)$ differ by an integer then we first obtain $y_1=\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r_1}$as a solution and then use reduction of order to find $y_2$. The formula for reduction for order is
$$y_2 = Cy_1\int\dfrac{e^{-\int p(x)dx}}{y_1^2(x)} dx$$
as derived here. Now, apparently you can derive the form of $y_2$ rigorously by substituting $y_1=\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r_1}$ in the above eqaution. However, I cant find this derivation anywhere. If anyone knows how to derive
$$y_2 = Cy_1\ln x + x^{r_2}\sum_{n=0}^{\infty}b_nx^n$$
using this method or if there is ANY source/textbook that does this, please let me know.
Best Answer
First, the formula of reduction of order is $y_2=\color\red{y_1}\int\frac{e^{-\int p(x)dx}}{y_1^2(x)}dx$.
Also note that we can ignore $C$ since the solution $c_1y_1+c_2y_2$ already contains general constants.
To derive the given formula, let's start with evaluating $\frac{1}{y_1^2(x)}=\frac{1}{x^{2r_1}(c_0+c_1x+c_2x^2+\cdots)}$.
Let $X=\frac{1}{c_0+c_1x+c_2x^2+\cdots}$, and observe that $X\cdot(c_0+c_1x+c_2x^2+\cdots)=1$.
By comparing the coefficient of each term, we can know that $X=\frac{1}{c_0}-\frac{c_1}{c_0^2}x+\frac{c_2c_0-c_1^2}{c_0^3}x^2+\cdots$.
So we can say that $\frac{1}{y_1^2(x)}=x^{-2r_1}(C_0+C_1x+C_2x^2+\cdots)$ for $C_0=\frac{1}{c_0},C_1=-\frac{c_1}{c_0^2},C_2=-\frac{c_2c_0-c_1^2}{c_0^3},\cdots$.
Now, observe that $p(x)=\frac{a_0}{x}+a_1+a_2x+\cdots$ so that $e^{-\int p(x)dx}=e^{-a_0\ln(x)-a_1x-\frac{a_1}{2}x^2\cdots}$
$=x^{-a_0}e^{-a_1x-\frac{a_1}{2}x^2\cdots}=x^{-a_0}(1+R(x)+\frac{1}{2}\{R(x)\}^2+\cdots)=x^{-a_0}(1+d_1x+d_2x^2+\cdots)$
where $R(x)=-a_1x-\frac{a_1}{2}x^2\cdots$, by using taylor series of $e^x$.
Thus we can write $\frac{e^{-\int p(x)dx}}{y_1^2(x)}=x^{-2r_1-a_0}(D_0+D_1x+D_2x^2+\cdots)$
$=D_0x^{-2r_1-a_0}+D_1x^{-2r_1-a_0+1}+\cdots+D_{2r_1+a_0-1}\frac{1}{x}+\cdots$.
Then $\int\frac{e^{-\int p(x)dx}}{y_1^2(x)}dx=D_{2r_1+a_0-1}\ln (x)+\left[\frac{D_0}{-2r_1-a_0+1}x^{-2r_1-a_0+1}+\frac{D_1}{-2r_1-a_0+2}x^{-2r_1-a_0+2}+\cdots\right]$.
Therefore $y_2=y_1(x)\ln (x)+x^{r_1}(E_0x^{-2r_1-a_0+1}+E_1x^{-2r_1-a_0+2}+\cdots)$.
Note that we divided RHS by $D_{2r_1+a_0-1}$.
Now, the proof ends by showing $-r_1-a_0+1\geq r_2$ and they differ by integer so that $(E_0x^{-r_1-a_0+1}+E_1x^{-r_1-a_0+2}+\cdots)$ can be written as $x^{r_2}\sum_{n=0}^\infty b_nx^n$.
Since $r_1$ and $r_2$ are the roots of the indicial equation $r^2-(1-a_0)r+b_0$, $r_1+r_2=1-a_0$.
Then $-r_1-a_0+1=r_2$, so they differ by an integer $0$, and the proof ends.
Moreover, we can know that $b_0 \neq 0$. (Your book may also mention this condition.)