1.3.11 Example 11 \(y^{\prime }+2xy=\sqrt {x}\)

\[ y^{\prime }+2xy=\sqrt {x}\]

Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) \) 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{align*} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+2x\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }2a_{n}x^{n+r+1} & =0 \end{align*}

Reindex so all powers on \(x\) are the lowest gives

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=2}^{\infty }2a_{n-2}x^{n+r-1}=0 \tag {1}\end{equation}

For \(n=0\,\), Eq(1) gives

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

Hence \(r=0\) since \(a_{0}\neq 0\). For \(n=1\), Eq(1) gives

\begin{align*} \left ( 1+r\right ) a_{1}x^{r} & =0\\ a_{1} & =0 \end{align*}

For \(n\geq 2\) the recurrence relation is from (1)

\begin{align} \left ( n+r\right ) a_{n}+2a_{n-2} & =0\nonumber \\ a_{n} & =-\frac {2a_{n-2}}{\left ( n+r\right ) } \tag {2}\end{align}

Or for \(r=0\) the above simplifies to

\begin{equation} a_{n}=-\frac {2}{n}a_{n-2} \tag {2A}\end{equation}

Eq (2A) is what is used to find all \(a_{n}\) for For \(n\geq 2\). Hence for \(n=2\) and remembering that \(a_{0}=1\) gives

\[ a_{2}=-1 \]

For \(n=3\)

\[ a_{3}=-\frac {2}{3}a_{1}=0 \]

For \(n=4\)

\[ a_{4}=-\frac {1}{2}a_{2}=\frac {1}{2}\]

For \(n=5,7,\cdots \) and all odd \(n\) then \(a_{n}=0\). For \(n=6\)

\[ a_{6}=-\frac {1}{3}a_{4}=-\frac {1}{6}\]

And so on. Hence (using \(a_{0}=1\))

\begin{align*} y_{h} & =c_{1}\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ & =c_{1}\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =c_{1}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \right ) \\ & =c_{1}\left ( 1-x^{2}+\frac {1}{2}x^{4}-\frac {1}{6}x^{6}+\cdots \right ) \end{align*}

Now we need to find \(y_{p}\) using the balance equation. From above we found that

\[ ra_{0}x^{r-1}=x^{\frac {1}{2}}\]

Renaming \(a\) to \(c\) the above becomes

\[ rc_{0}x^{r-1}=x^{\frac {1}{2}}\]

Hence \(r-1=\frac {1}{2}\) or \(r=\frac {3}{2}\). Therefore \(rc_{0}=1\) or \(c_{0}=\frac {2}{3}\). To find the rest of \(c_{n}\), we see that summation terms in (1) above, do not all have same starting index, then here, we can not just find \(c_{0}\) and assume that all \(c_{n}=0\) for \(n>0\) as we did in the earlier example. In this case we have to use the same relation (1) above to find all the \(c_{n}\) like we did when finding \(a_{n}\). We just have to use \(c_{n}\) for \(a_{n}\) and use \(r\) found from the balance equation (*). EQ (1) becomes

\begin{align} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=2}^{\infty }2a_{n-2}x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+\frac {3}{2}\right ) c_{n}x^{n+\frac {3}{2}-1}+\sum _{n=2}^{\infty }2c_{n-2}x^{n+\frac {3}{2}-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+\frac {3}{2}\right ) c_{n}x^{n+\frac {1}{2}}+\sum _{n=2}^{\infty }2c_{n-2}x^{n+\frac {1}{2}} & =0 \tag {1B}\end{align}

And it is (1B) that will be used to find all \(c_{n}\) for \(n>0\). For \(n=1\)

\begin{align*} \left ( n+\frac {3}{2}\right ) c_{n} & =0\\ \left ( 1+\frac {3}{2}\right ) c_{1} & =0\\ c_{1} & =0 \end{align*}

For \(n=2\) we have recursion relation

\begin{align*} \left ( n+\frac {3}{2}\right ) c_{n}+2c_{n-2} & =0\\ c_{n} & =\frac {-2c_{n-2}}{n+\frac {3}{2}}\end{align*}

Hence for \(n=2\)

\begin{align*} c_{2} & =\frac {-2c_{0}}{\left ( 2+\frac {3}{2}\right ) }\\ & =\frac {-2\left ( \frac {2}{3}\right ) }{\left ( 2+\frac {3}{2}\right ) }\\ & =-\frac {8}{21}\end{align*}

For \(n=3\)

\begin{align*} c_{3} & =\frac {-2c_{1}}{\left ( 3+\frac {3}{2}\right ) }\\ & =0 \end{align*}

For \(n=4\)

\begin{align*} c_{4} & =\frac {-2c_{2}}{\left ( 4+\frac {3}{2}\right ) }\\ & =\frac {-2\left ( -\frac {8}{21}\right ) }{\left ( 4+\frac {3}{2}\right ) }\\ & =\frac {32}{231}\end{align*}

And so on. Hence

\begin{align*} y_{p} & =x^{\frac {3}{2}}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {3}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \end{align*}

Hence the final solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-x^{2}+\frac {1}{2}x^{4}-\frac {1}{6}x^{6}+\cdots \right ) +x^{\frac {3}{2}}\left ( \frac {2}{3}+-\frac {8}{21}x^{2}+\frac {32}{231}x^{4}-\cdots \right ) \end{align*}