3.3.16.1 Example 1

\begin{align} y^{\prime }+y\cot x & =y^{4}\tag {1}\\ y\left ( 0\right ) & =0\nonumber \end{align}

Comparing to \(y^{\prime }+py=qy^{n}\) shows that \(p=\cot x,q=1,n=4\). Let \(v=y^{1-n}=y^{1-4}=y^{-3}\).  Then \(\frac {dv}{dx}=-3y^{-4}y^{\prime }\) or \(y^{\prime }=\frac {v^{\prime }}{-3y^{-4}}\). The ode becomes

\[ \frac {v^{\prime }}{-3y^{-4}}+y\cot x=y^{4}\]

Multiplying both sides by \(y^{-4}\) gives

\[ \frac {v^{\prime }}{-3}+y^{-3}\cot x=1 \]

But \(y^{-3}=v\) and the above becomes

\begin{align*} \frac {v^{\prime }}{-3}+v\cot x & =1\\ v^{\prime }-3v\cot x & =-3 \end{align*}

Which is linear in \(v\).  Solving gives

\begin{align*} v & =\frac {1}{4}\left ( 3\sin x-\sin \left ( 3x\right ) \right ) \left ( \frac {3}{2}\csc x\cot x-\frac {3}{2}\ln \left ( \csc \left ( x\right ) -\cot x\right ) +c_{1}\right ) \\ & =\left ( \sin x\right ) ^{3}\left ( \frac {3}{2}\csc x\cot x-\frac {3}{2}\ln \left ( \csc \left ( x\right ) -\cot x\right ) +c_{1}\right ) \end{align*}

But \(v=\frac {1}{y^{3}}\). Hence the solution is

\[ \frac {1}{y^{3}}=\left ( \sin x\right ) ^{3}\left ( \frac {3}{2}\csc x\cot x-\frac {3}{2}\ln \left ( \csc \left ( x\right ) -\cot x\right ) +c_{1}\right ) \]

Was not able to solve for \(c_{1}\) at the given IC since gives 1/0.  Hence only trivial solution exist, which is

\[ y=0 \]