1.2.3.1 Algorithm

In this case the solution is

\[ y=c_{1}y_{1}+c_{2}y_{2}\]

There are two sub cases that show up when roots differ by integer. First sub case is when the second solution \(y_{2}\) is obtained similar to how \(y_{1}\) is obtained. i.e. using standard Frobenius series but with the second root. The second sub case is the harder one, this is when \(y_{2}\) fails to be obtained using the standard method due to \(b_{N}\) being undefined where \(N\) is the difference between the roots. In this sub case we need to use a modified Frobenius series method where, which is explained more using examples below. Therefore for sub case one (called the good case) we have

\begin{align*} y_{1} & =x^{r_{1}}\sum _{n=0}^{\infty }a_{n}x^{n}\\ y_{2} & =Cy_{1}\ln \left ( x\right ) +x^{r_{2}}\sum _{n=0}^{\infty }b_{n}x^{n}\end{align*}

Where \(C\) above come out to be zero (good case). Hence the above simplifies to

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

For the second subcase (called the bad case), \(C\) above is not zero. To determine if \(C=0\) or not, we first find all the \(a_{n}\), including \(a_{N}\) where \(N\) is difference between the roots. Then evaluate

\[ \lim _{r\rightarrow r_{2}}a_{N}\left ( r\right ) \]

And if this exists, then \(C=0\). In the above we have to keep \(a_{n}\left ( r\right ) \) as function of \(r\) in symbolic form to do this.