4.9.1.1.8 Example 8 \(\left ( 1+\left ( y^{\prime }\right ) ^{2}\right ) ^{2}=y^{2}y^{\prime \prime },y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =\sqrt {2}\)

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

With IC

\begin{align*} y\left ( 0\right ) & =3\\ y^{\prime }\left ( 0\right ) & =\sqrt {2}\end{align*}

Let \(p=y^{\prime }\), hence \(y^{\prime \prime }=pp^{\prime }\) and the ode becomes

\begin{align} \left ( 1+p^{2}\right ) ^{2} & =y^{2}pp^{\prime }\nonumber \\ \frac {pp^{\prime }}{\left ( 1+p^{2}\right ) ^{2}} & =\frac {1}{y^{2}} \tag {2}\end{align}

Solving the above ode gives

\begin{align} p_{1} & =-\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) }\tag {3}\\ p_{2} & =\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) } \tag {4}\end{align}

Now we replace back \(p=y^{\prime }\left ( x\right ) \) above gives

\begin{align} y^{\prime } & =-\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) }\tag {3A}\\ y^{\prime } & =\frac {\sqrt {-2\left ( c_{1}y-1\right ) \left ( 2c_{1}y+y-2\right ) }}{2\left ( c_{1}y-1\right ) } \tag {4A}\end{align}

Lets start with (3A). Before solving, we will get rid of \(c_{1}\) using IC. Given that \(y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =\sqrt {2}\) then (3A) becomes

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

Hence (4A) becomes

\[ y^{\prime }=-\frac {\sqrt {-2\left ( \frac {1}{6}y-1\right ) \left ( 3y+y-2\right ) }}{2\left ( \frac {1}{6}y-1\right ) }\]

Solving this ode gives the solution

\begin{equation} x-\frac {\sqrt {-4y^{2}+30y-36}}{4}-\frac {9\arcsin \left ( \frac {4y}{9}-\frac {5}{3}\right ) }{8}+c_{2}=0 \tag {5}\end{equation}

Finally, using \(y\left ( 0\right ) =3\) the above becomes

\[ -\frac {\sqrt {-4\left ( 9\right ) +30\left ( 3\right ) -36}}{4}-\frac {9\arcsin \left ( \frac {4\left ( 3\right ) }{9}-\frac {5}{3}\right ) }{8}+c_{2}=0 \]

Solving for \(c_{2}\) gives

\[ c_{2}=\frac {\sqrt {18}}{4}-\frac {9\arcsin \left ( \frac {1}{3}\right ) }{8}\]

Hence (5) becomes

\begin{equation} x-\frac {\sqrt {-4y^{2}+30y-36}}{4}-\frac {9\arcsin \left ( \frac {4y}{9}-\frac {5}{3}\right ) }{8}+\frac {\sqrt {18}}{4}-\frac {9\arcsin \left ( \frac {1}{3}\right ) }{8}=0 \tag {6}\end{equation}

Now we have to do the same for ode (4A).  Given that \(y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =\sqrt {2}\) then (4A) becomes

\[ \sqrt {2}=\frac {\sqrt {-2\left ( 3c_{1}-1\right ) \left ( 6c_{1}+3-2\right ) }}{2\left ( 3c_{1}-1\right ) }\]

But there is no solution for \(c_{1}\). This means (4A) leads to no solution. Hence only solution is (6).