3.3.5.5 Example 5
\begin{align*} y^{\prime }-y\cot \left ( x\right ) & =-\frac {\sin x}{x^{2}}\\ y\left ( \infty \right ) & =0 \end{align*}
In normal form the ode is
\[ y^{\prime }+p\left ( x\right ) y=q\left ( x\right ) \]
Hence \(p\left ( x\right ) =-\cot \left ( x\right ) \) and \(q\left ( x\right ) =-\frac {\sin x}{x^{2}}\). Not deļ¬ned at IC, hence then uniqueness and
existence theory do not apply. The integrating factor is
\begin{align*} \mu & =e^{\int p\left ( x\right ) dx}\\ & =e^{\int -\cot \left ( x\right ) dx}\\ & =e^{-\ln \left ( \sin x\right ) }\\ & =\frac {1}{\sin x}\end{align*}
Then the ode becomes
\begin{align*} \frac {d}{dx}\left ( y\mu \right ) & =\mu q\left ( x\right ) \\ \frac {d}{dx}\left ( y\frac {1}{\sin x}\right ) & =-\frac {1}{\sin x}\left ( \frac {\sin x}{x^{2}}\right ) \\ \frac {y}{\sin x} & =-\int \frac {1}{x^{2}}\ dx+c\\ \frac {y}{\sin x} & =\frac {1}{x}+c\\ y & =\frac {\sin x}{x}+c\sin x\\ & =\sin \left ( x\right ) \left ( \frac {1}{x}+c\right ) \end{align*}
Applying IC gives
\[ 0=\sin \left ( x\right ) \left ( \frac {1}{x}+c\right ) \]
Either \(\sin x=0\) or \(\left ( \frac {1}{x}+c\right ) =0\). We look only at second equation, since that one has the \(c\) in it
which we want to solve. hence
\[ \left ( \frac {1}{x}+c\right ) =0 \]
As \(x\rightarrow \infty \) then \(\frac {1}{x}\rightarrow 0\) and we obtain \(c=0\). Hence the solution is
\[ y=\frac {\sin \left ( x\right ) }{x}\]