4.9.2.1.7 Example 6 \(2\left ( y^{\prime }\right ) ^{2}y^{\prime \prime }=-1\)
\begin{align} y^{\prime \prime } & =-\frac {1}{2\left ( y^{\prime }\right ) ^{2}}\tag {1}\\ y\left ( 0\right ) & =1\nonumber \\ y^{\prime }\left ( 0\right ) & =-1\nonumber \end{align}
Notice that this is also missing \(x\) type second order ode. Now let \(p\left ( x\right ) =y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\) and the ode
becomes
\[ p^{\prime }=-\frac {1}{2p^{2}}\]
Which is quadrature. The solution is
\[ p^{3}+\frac {3x}{2}=c_{1}\]
At
\(x=0,p\left ( 0\right ) =-1\). Hence the above gives
\[ -1=c_{1}\]
And the solution
becomes
\[ p^{3}+\frac {3x}{2}=-1 \]
But
\(p=y^{\prime }\), hence the above becomes
\[ \left ( y^{\prime }\right ) ^{3}+\frac {3x}{2}=-1 \]
With IC
\(y\left ( 0\right ) =1\). This is quadrature. Solving
gives
\begin{align*} y_{1} & =-\frac {1}{16}i\left ( 3x+2\right ) \left ( -i+\sqrt {3}\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+c_{1}\\ y_{2} & =\frac {1}{16}i\left ( 3x+2\right ) \left ( i+\sqrt {3}\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+c_{1}\\ y_{3} & =\frac {1}{8}\left ( 3x+2\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+c_{1}\end{align*}
Applying IC to the above shows that only second solution satisfies the original initial
conditions with \(c=\frac {3}{2}\). Hence solution is
\[ y_{2}=\frac {1}{16}\left ( 3x+2\right ) \left ( i\sqrt {3}-1\right ) \left ( -12x-8\right ) ^{\frac {1}{3}}+\frac {3}{2}\]
Another option when solving these types of odes is not
to plugin the IC until the very end. Like this. Starting with
\[ p^{3}+\frac {3x}{2}=c_{1}\]
We do not resolve
the
\(c_{1}\). But keep it. Since
\(p=y^{\prime }\), hence the above becomes
\[ \left ( y^{\prime }\right ) ^{3}+\frac {3x}{2}=c_{1}\]
This is quadrature. Solving
gives
\begin{align*} y_{1} & =\frac {1}{16}i\left ( 3x-2c_{1}\right ) \left ( i-\sqrt {3}\right ) \left ( 8c_{1}-12x\right ) ^{\frac {1}{3}}+c_{2}\\ y_{2} & =\frac {1}{16}i\left ( 3x-2c_{1}\right ) \left ( i+\sqrt {3}\right ) \left ( 8c_{1}-12x\right ) ^{\frac {1}{3}}+c_{2}\\ y_{3} & =\frac {1}{8}\left ( 3x-2c_{1}\right ) \left ( 8c_{1}-12x\right ) ^{\frac {1}{3}}+c_{2}\end{align*}
And only now we solve for \(c_{1},c_{2}\) from both initial conditions. Assuming we made no mistake,
then same result will come out.