1.3.16 Example 16 \(\cos \left ( x\right ) y^{\prime }+\frac {1}{x}y=x\)

\[ \cos \left ( x\right ) y^{\prime }+\frac {1}{x}y=x \]

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}

The homogeneous ode becomes

\begin{align*} \cos \left ( x\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+x^{-1}\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r-1} & =0\\ \left ( \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-\frac {1}{2}x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\frac {1}{24}x^{4}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\cdots \right ) +\sum _{n=0}^{\infty }a_{n}x^{n+r-1} & =0\\ \left ( \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\frac {1}{2}\left ( n+r\right ) a_{n}x^{n+r+1}+\sum _{n=0}^{\infty }\frac {1}{24}\left ( n+r\right ) a_{n}x^{n+r+3}+\cdots \right ) +\sum _{n=0}^{\infty }a_{n}x^{n+r-1} & =0 \end{align*}

Making all powers on \(x\) the lowest, which is \(n+r-1\) gives

\begin{equation} \left ( \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {1}{2}\left ( n+r-2\right ) a_{n-2}x^{n+r-1}+\sum _{n=4}^{\infty }\frac {1}{24}\left ( n+r-4\right ) a_{n-4}x^{n+r-1}+\cdots \right ) +\sum _{n=0}^{\infty }a_{n}x^{n+r-1}=0 \tag {1}\end{equation}

The indicial equation is when \(n=0\). Hence

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

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

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

Hence \(a_{1}=0\).  For \(n=2\)

\begin{align*} \left ( n+r\right ) a_{n}x^{n+r-1}-\frac {1}{2}\left ( n+r-2\right ) a_{n-2}x^{n+r-1}+a_{n}x^{n+r-1} & =0\\ \left ( 2-1\right ) a_{2}-\frac {1}{2}\left ( 2-1-2\right ) a_{2-2}+a_{2} & =0\\ a_{2}+\frac {1}{2}a_{0}+a_{2} & =0\\ 2a_{2} & =-\frac {1}{2}a_{0}\\ a_{2} & =-\frac {1}{4}a_{0}\end{align*}

For \(n=3\)

\begin{align*} \left ( n+r\right ) a_{n}x^{n+r-1}-\frac {1}{2}\left ( n+r-2\right ) a_{n-2}x^{n+r-1}+a_{n}x^{n+r-1} & =0\\ \left ( 3-1\right ) a_{3}-\frac {1}{2}\left ( 3-1-2\right ) a_{3-2}+a_{3} & =0\\ 2a_{3}+a_{3} & =0\\ a_{3} & =0 \end{align*}

For \(n=4\)

\begin{align*} \left ( n+r\right ) a_{n}x^{n+r-1}-\frac {1}{2}\left ( n+r-2\right ) a_{n-2}x^{n+r-1}+\frac {1}{24}\left ( n+r-4\right ) a_{n-4}x^{n+r-1}+a_{n}x^{n+r-1} & =0\\ \left ( 4-1\right ) a_{4}-\frac {1}{2}\left ( 4-1-2\right ) a_{2}+\frac {1}{24}\left ( 4-1-4\right ) a_{0}+a_{4} & =0\\ 3a_{4}-\frac {1}{2}a_{2}-\frac {1}{24}a_{0}+a_{4} & =0\\ 4a_{4} & =\frac {1}{2}a_{2}+\frac {1}{24}a_{0}\\ 4a_{4} & =\frac {1}{2}\left ( -\frac {1}{4}a_{0}\right ) +\frac {1}{24}a_{0}\\ & =-\frac {1}{8}a_{0}+\frac {1}{24}a_{0}\\ & =-\frac {1}{12}a_{0}\\ a_{4} & =-\frac {1}{48}a_{0}\end{align*}

And so on. Hence

\begin{align*} y_{h} & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ & =\sum _{n=0}^{\infty }a_{n}x^{n-1}\\ & =\frac {1}{x}\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =\frac {1}{x}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =\frac {1}{x}\left ( a_{0}+0x-\frac {1}{4}a_{0}x^{2}+0x^{3}-\frac {1}{48}a_{0}x^{4}+\cdots \right ) \\ & =\frac {1}{x}\left ( a_{0}-\frac {1}{4}a_{0}x^{2}-\frac {1}{48}a_{0}x^{4}+\cdots \right ) \\ & =a_{0}\left ( \frac {1}{x}-\frac {1}{4}x-\frac {1}{48}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)

\[ c_{0}\left ( r+1\right ) x^{r-1}=x \]

For balance we need \(r-1=1\) or \(r=2\). Hence \(c_{0}\left ( r+1\right ) =1\) or \(c_{0}=\frac {1}{3}\). 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 \(r=2\) instead of \(r=-1\). EQ (1) now becomes

\begin{align} \left ( \sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {1}{2}\left ( n+r-2\right ) c_{n-2}x^{n+r-1}+\sum _{n=4}^{\infty }\frac {1}{24}\left ( n+r-4\right ) c_{n-4}x^{n+r-1}+\cdots \right ) +\sum _{n=0}^{\infty }c_{n}x^{n+r-1} & =0\nonumber \\ \left ( \sum _{n=0}^{\infty }\left ( n+2\right ) c_{n}x^{n+1}-\sum _{n=2}^{\infty }\frac {1}{2}nc_{n-2}x^{n+1}+\sum _{n=4}^{\infty }\frac {1}{24}\left ( n-2\right ) c_{n-4}x^{n+1}+\cdots \right ) +\sum _{n=0}^{\infty }c_{n}x^{n+1} & =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+2\right ) c_{n}+c_{n} & =0\\ 3c_{1}+c_{1} & =0\\ c_{1} & =0 \end{align*}

For \(n=2\)

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

Since \(c_{0}=\frac {1}{3}\). For \(n=3\)

\begin{align*} \left ( n+2\right ) c_{n}-\frac {1}{2}nc_{n-2}+c_{n} & =0\\ 5c_{3}-\frac {3}{2}c_{1}+c_{3} & =0\\ 6c_{3} & =0\\ c_{3} & =0 \end{align*}

For \(n=4\)

\begin{align*} \left ( n+2\right ) c_{n}-\frac {1}{2}nc_{n-2}+\frac {1}{24}\left ( n-2\right ) c_{n-4}+c_{n} & =0\\ 6c_{4}-2c_{2}+\frac {1}{12}c_{0}+c_{4} & =0\\ 7c_{4} & =2c_{2}-\frac {1}{12}c_{0}\\ & =2\left ( \frac {1}{15}\right ) -\frac {1}{12}\left ( \frac {1}{3}\right ) \\ & =\frac {19}{180}\\ c_{4} & =\frac {19}{180\left ( 7\right ) }\\ & =\frac {19}{1260}\end{align*}

And so on. Hence

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+r}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n+2}\\ & =x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{3}+0x+\frac {1}{15}x^{2}+0x^{3}+\frac {19}{1260}x^{4}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{3}+\frac {1}{15}x^{2}+\frac {19}{1260}x^{4}+\cdots \right ) \\ & =\frac {x^{2}}{3}+\frac {x^{4}}{15}+\frac {19}{1260}x^{6}+\cdots \end{align*}

Hence the solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =a_{0}\left ( \frac {1}{x}-\frac {1}{4}x-\frac {1}{48}x^{3}+\cdots \right ) +\left ( \frac {x^{2}}{3}+\frac {x^{4}}{15}+\frac {19}{1260}x^{6}+\cdots \right ) \\ & =c_{1}\left ( \frac {1}{x}-\frac {1}{4}x-\frac {1}{48}x^{3}+\cdots \right ) +\left ( \frac {x^{2}}{3}+\frac {x^{4}}{15}+\frac {19}{1260}x^{6}+\cdots \right ) \end{align*}