4.9.1.1.3 Example 3 \(2yy^{\prime \prime }-y^{3}-2\left ( y^{\prime }\right ) ^{2}=0,y\left ( 0\right ) =-1,y^{\prime }\left ( 0\right ) =0\)

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

Before solving this ode, we can either use initial conditions to solve for \(c_{1}\) or solve it as it is and at the very end use initial conditions to solve for both \(c_{1}\) and the new constant which will come up which will be \(c_{2}\).  It is easier to get rid of \(c_{1}\) now than keep it. Will show both methods.

Getting rid of \(c_{1}\) now method. 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*}

Eq(3A) 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\nonumber \\ \operatorname {arctanh}\left ( \sqrt {y+1}\right ) & =-\frac {x}{2}\nonumber \\ \sqrt {y+1} & =\tanh \left ( -\frac {x}{2}\right ) \nonumber \\ & =-\tanh \left ( \frac {x}{2}\right ) \nonumber \\ y+1 & =\tanh ^{2}\left ( \frac {x}{2}\right ) \nonumber \\ y & =\tanh ^{2}\left ( \frac {x}{2}\right ) -1 \tag {5}\end{align}

Now we solve the second ode (4A). At \(x=0\) we have \(y^{\prime }\left ( 0\right ) =0,y\left ( 0\right ) =-1\) hence Eq.(4A) becomes

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

Hence (4A) becomes

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

Which gives the solution

\begin{equation} y\left ( x\right ) =x+2\arctan \left ( y-1\right ) +c_{2} \tag {6}\end{equation}

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

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

Hence the solution (6) becomes

\[ y\left ( x\right ) =x+2\arctan \left ( y-1\right ) -1+2\arctan \left ( 2\right ) \]

But this solution does not satisfy \(y^{\prime }\left ( 0\right ) =0\). Hence it is not valid solution. So the only solution is (5).

Now we will do the same thing, but we will not get rid of \(c_{1}\) early one as above, and keep it until the end. We will see we will get same solution as (5).

Not getting rid of \(c_{1}\) method. Starting from (3A) and (4A) above.

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

Starting with (3A), solving it gives

\begin{align} \int \frac {1}{\sqrt {y+c_{1}}y}dy & =x+c_{2}\nonumber \\ -\frac {2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) }{\sqrt {c_{1}}} & =x+c_{2}\nonumber \\ -2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =x\sqrt {c_{1}}+c_{3}\nonumber \\ \operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =-x\frac {\sqrt {c_{1}}}{2}+c_{4}\nonumber \\ \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}} & =\tanh \left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ \sqrt {y+c_{1}} & =\sqrt {c_{1}}\tanh \left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y+c_{1} & =c_{1}\tanh ^{2}\left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y & =c_{1}\tanh ^{2}\left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) -c_{1} \tag {7}\end{align}

Now we can solve for the initial conditions. using \(y\left ( 0\right ) =-1\) gives

\begin{equation} -1=c_{1}\tanh ^{2}\left ( c_{4}\right ) -c_{1} \tag {8}\end{equation}

Taking derivative of the above gives

\[ y^{\prime }=-c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) \operatorname {sech}\left ( c_{4}-x\frac {\sqrt {c_{1}}}{2}\right ) ^{2}\]

Applying \(y^{\prime }\left ( 0\right ) =0\) gives

\begin{equation} 0=-c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}\right ) \operatorname {sech}\left ( c_{4}\right ) ^{2} \tag {9}\end{equation}

Solving (8,9) for \(c_{1},c_{4}\) gives

\begin{align*} c_{1} & =1\\ c_{4} & =0 \end{align*}

Hence the solution (7) is

\begin{equation} y=\tanh ^{2}\left ( -\frac {1}{2}x\right ) -1 \tag {10}\end{equation}

Which same as (5).  Now we go back and solve (4A).

\begin{align} y^{\prime }\left ( x\right ) & =-y\sqrt {y+c_{1}}\nonumber \\ \int \frac {1}{\sqrt {y+c_{1}}y}dy & =-x+c_{2}\\ -\frac {2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) }{\sqrt {c_{1}}} & =-x+c_{2}\nonumber \\ -2\operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =-x\sqrt {c_{1}}+c_{3}\nonumber \\ \operatorname {arctanh}\left ( \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}}\right ) & =x\frac {\sqrt {c_{1}}}{2}+c_{4}\nonumber \\ \frac {\sqrt {y+c_{1}}}{\sqrt {c_{1}}} & =\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ \sqrt {y+c_{1}} & =\sqrt {c_{1}}\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y+c_{1} & =c_{1}\tanh ^{2}\left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \nonumber \\ y & =c_{1}\tanh ^{2}\left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) -c_{1} \tag {7}\end{align}

Now we can solve for the initial conditions. using \(y\left ( 0\right ) =-1\) gives

\begin{equation} -1=c_{1}\tanh ^{2}\left ( c_{4}\right ) -c_{1} \tag {8}\end{equation}

Taking derivative of (7) gives

\[ y^{\prime }=c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) \left ( 1-\tanh \left ( c_{4}+x\frac {\sqrt {c_{1}}}{2}\right ) ^{2}\right ) \]

Applying \(y^{\prime }\left ( 0\right ) =0\) gives

\begin{equation} 0=c_{1}^{\frac {3}{2}}\tanh \left ( c_{4}\right ) \left ( 1-\tanh \left ( c_{4}\right ) ^{2}\right ) \tag {9}\end{equation}

But now if we try to solve (8,9) for \(c_{1},c_{4}\) we see no solution exists. Hence (4A) leads to no solution. Only solution is (8). This is the same as earlier method.

This shows that if we get rid of \(c_{1}\) early one or not, same solution results. But it is much easier to get rid of \(c_{1}\) after finding the solution to the first ode.