1.3.13 Example 13 \(y^{\prime }=\frac {1}{x}\)

\[ y^{\prime }=\frac {1}{x}\]

Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) =0\) is analytic at \(x=0\). However the RHS has no series expansion at \(x=0\) (not analytic there). Therefore we must use Frobenius series in this case.  Let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\end{align*}

The (homogenous) ode becomes

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}=0 \tag {1}\end{equation}

For \(n=0\)

\begin{equation} ra_{0}x^{r-1}=0 \tag {*}\end{equation}

Hence \(r=0\) since \(a_{0}\neq 0\). Therefore the ode satisfies

\[ y^{\prime }=ra_{0}x^{r-1}\]

Eq (1) becomes

\begin{align} \sum _{n=0}^{\infty }na_{n}x^{n-1} & =0\nonumber \\ na_{n}x^{n-1} & =0 \tag {2}\end{align}

Therefore for all \(n\geq 1\) we have \(a_{n}=0\). Hence

\[ y_{h}=a_{0}\]

Now we need to find \(y_{p}\) using the balance equation (*). From above we see that (where we rename \(a_{0}\) to \(c_{0}\))

\[ rc_{0}x^{r-1}=x^{-1}\]

Hence \(r-1=-1\) or \(r=0\). Hence

\begin{align*} rc_{0} & =1\\ 0c_{0} & =1 \end{align*}

Therefore there is no solution for \(c_{0}\). Unable to find \(y_{p}\) therefore no series solution exists. Asymptotic methods are needed to solve this. Mathematica AsymptoticDSolveValue gives the solution as \(y\left ( x\right ) =c+\ln x\).