Example 17 xy′ + 2xy = √

Expansion is around \(x=0\). Since regular singular point, then we must use Frobenius series in this case. Let

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

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

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

\begin{align} \left ( n+r\right ) a_{n}x^{n+r} & =0\nonumber \\ ra_{0}x^{r} & =0\nonumber \\ ra_{0} & =0 \tag {*}\end{align}

Hence \(r=0\) since \(a_{0}\neq 0\). EQ (*) is the balance equation. Now we ﬁnd all \(a_{n}\). For \(n=1\), EQ (1) gives (and using \(r=0\))

\begin{align*} a_{1}+2a_{0} & =0\\ a_{1} & =-2a_{0}\end{align*}

\begin{align*} 2a_{2}+2a_{1} & =0\\ 2a_{2} & =-2a_{1}\\ a_{2} & =-a_{1}\\ & =2a_{0}\end{align*}

\begin{align*} 3a_{3}+2a_{2} & =0\\ a_{3} & =-\frac {2a_{2}}{3}\\ a_{3} & =-\frac {4a_{0}}{3}\end{align*}

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

Now that we found \(y_{h}\), we need to ﬁnd \(y_{p}\). The balance equation is (*). Using \(c_{0}\) instead of \(a_{0}\) it becomes (now with \(x\) on the right side)

For balance we need \(r=\frac {1}{2}\). Hence \(c_{0}r=1\) or \(c_{0}=2\). Since the summation terms in (1) do not all have the same starting index, then we have to use (1) to ﬁnd all \(c_{n}\) and we can not just use \(c_{0}\) like we did in earlier problem. Only when starting index is the same on all summation terms, we can do that.

To ﬁnd all \(c_{n}\) we repeat the same process used to ﬁnd \(a_{n}\). We use EQ (1) again but replace all \(a_{n}\) by \(c_{n}\) and now use \(r=\frac {1}{2}\) instead of \(r=0\). EQ (1) now becomes

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

Notice that we kept the right side zero, since we used \(n=0\) for balance and hence there is no more terms to balance any more for \(n>0\). Now we use (1B) to ﬁnd all \(c_{n}\) for \(n>0\). For \(n=1\)

\begin{align*} \left ( n+\frac {1}{2}\right ) c_{n}+2c_{n-1} & =0\\ \frac {3}{2}c_{1}+2c_{0} & =0\\ c_{1} & =-\frac {4}{3}c_{0}\\ & =-\frac {4}{3}2\\ & =-\frac {8}{3}\end{align*}

\begin{align*} \left ( 2+\frac {1}{2}\right ) c_{2}+2c_{1} & =0\\ \frac {5}{2}c_{2} & =-2c_{1}\\ c_{2} & =-\frac {4c_{1}}{5}\\ c_{2} & =\left ( -\frac {4}{5}\right ) \left ( -\frac {8}{3}\right ) \\ & =\frac {32}{15}\end{align*}

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+r}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n+\frac {1}{2}}\\ & =\sqrt {x}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =\sqrt {x}\left ( c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \right ) \\ & =\sqrt {x}\left ( 2-\frac {8}{3}x+\frac {32}{15}x^{2}-\frac {128}{105}x^{3}+\frac {512}{945}x^{4}-\cdots \right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =a_{0}\left ( 1-2x+2x^{2}-\frac {4}{3}x^{3}+\cdots \right ) +\sqrt {x}\left ( 2-\frac {8}{3}x+\frac {32}{15}x^{2}-\frac {128}{105}x^{3}+\frac {512}{945}x^{4}-\cdots \right ) \end{align*}

1.3.17 Example 17 \(xy^{\prime }+2xy=\sqrt {x}\)