4.9.1.1.6 Example 6 \(2y^{\prime \prime }-\sin \left ( 2y\right ) =0,y\left ( 0\right ) =-\frac {\pi }{2},y^{\prime }\left ( 0\right ) =1\)

\begin{equation} 2y^{\prime \prime }-\sin \left ( 2y\right ) =0 \tag {1}\end{equation}

With IC

\begin{align*} y\left ( 0\right ) & =-\frac {\pi }{2}\\ y^{\prime }\left ( 0\right ) & =1 \end{align*}

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes

\begin{align} 2p\frac {dp}{dy} & =\sin \left ( 2y\right ) \tag {2}\\ 2pdp & =\sin \left ( 2y\right ) dy\nonumber \\ \int 2pdp & =\int \sin \left ( 2y\right ) dy\nonumber \\ p^{2} & =-\frac {1}{2}\cos \left ( 2y\right ) +c_{1}\nonumber \end{align}

At \(x=0\) we have \(p=1,y=-\frac {\pi }{2}\). Hence the above becomes

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

Therefore (2) becomes

\[ \left ( y^{\prime }\left ( x\right ) \right ) ^{2}=-\frac {1}{2}\cos \left ( 2y\right ) +\frac {1}{2}\]

Need to solve and apply IC \(y\left ( 0\right ) =-\frac {\pi }{2}\) to finish.