Example 3 \begin {equation} 2yy^{\prime \prime }-y^{3}-2\left ( y^{\prime }\right ) ^{2}=0 \tag {1} \end {equation} With IC\begin {align*} y\left ( 0\right ) & =-1\\ y^{\prime }\left ( 0\right ) & =0 \end {align*} Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes\begin {align} 2yp\frac {dp}{dy}-y^{3}-2p^{2} & =0\tag {2}\\ \frac {dp}{dy} & =\frac {y^{3}+2p^{2}}{2py}\nonumber \end {align}

Which is first order ode in \(p\left ( y\right ) \) of type Bernoulli. There are two solutions\begin {align} p_{1} & =y\sqrt {y+c_{1}}\tag {3}\\ p_{2} & =-y\sqrt {y+c_{1}} \tag {4} \end {align}

But \(p=y^{\prime }\) hence the above becomes\begin {align} y^{\prime }\left ( x\right ) & =y\sqrt {y+c_{1}}\tag {3}\\ y^{\prime }\left ( x\right ) & =-y\sqrt {y+c_{1}} \tag {4} \end {align}

Solving (3). At \(x=0\) we have \(y^{\prime }\left ( 0\right ) =0,y\left ( 0\right ) =-1\) hence the above becomes

\begin {align*} 0 & =-1\sqrt {-1+c_{1}}\\ 0 & =\sqrt {-1+c_{1}}\\ c_{1} & =1 \end {align*}

Hence (3) becomes

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

This is quadrature. Integrating

\begin {align*} \frac {dy}{y\sqrt {y+1}} & =dx\\ -2\operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =x+c_{2} \end {align*}

At \(x=0\) we have\(\ y\left ( 0\right ) =-1\) and the above becomes

\begin {align*} -2\operatorname {arctanh}\left ( \sqrt {-1+1}\right ) & =c_{2}\\ c_{2} & =-2\operatorname {arctanh}\left ( 0\right ) \\ c_{2} & =0 \end {align*}

Hence the solution is

\begin {align*} -2\operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =x\\ \operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =-\frac {x}{2}\\ \sqrt {y+1} & =\tanh \left ( -\frac {x}{2}\right ) \\ & =-\tanh \left ( \frac {x}{2}\right ) \\ y+1 & =\tanh ^{2}\left ( \frac {x}{2}\right ) \\ y & =\tanh ^{2}\left ( \frac {x}{2}\right ) -1 \end {align*}

Solving (4)

At \(x=0\) we have \(y^{\prime }\left ( 0\right ) =0,y\left ( 0\right ) =-1\) hence

\begin {align*} 0 & =1\sqrt {-1+c_{1}}\\ 0 & =\sqrt {-1+c_{1}}\\ c_{1} & =1 \end {align*}

Hence (4) becomes

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

Which gives same solution as before. \(y=\tanh ^{2}\left ( \frac {x}{2}\right ) -1\)