2.5.1 Algorithm

If one of the roots is an integer, and the ode is inhomogeneous. ode, then we do not need to split the solution into \(y_{h},y_{p}\) and can use the integer root to find \(y_{p}\) directly. If both roots are non-integer, we have to split the problem into \(y_{h},y_{p}\). This is because it will not be possible to match powers on \(x\) from the left side to the right side. Because the RHS will be polynomial in \(x\), but the LHS will not be polynomial in \(x\) because of the non integer powers on \(x.\)In this case the solution is

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

Where

\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*}

And \(r_{1},r_{2}\) are roots of the indicial equation. \(a_{0},b_{0}\) are set to \(1\) as arbitrary.