4.9.1.1.5 Example 5 \(2y^{\prime \prime }-e^{y}=0,y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =1\)

This is same example as above, but here we delay applying IC to the very end to see the difference. This method is more general, but makes solving for IC harder.

\begin{equation} 2y^{\prime \prime }-e^{y}=0 \tag {1}\end{equation}

With IC

\begin{align*} y\left ( 0\right ) & =0\\ 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{equation} 2\frac {dp}{dy}p=e^{y}\nonumber \end{equation}

This is separable.

\begin{align*} 2\int pdp & =\int e^{y}dy\\ p^{2} & =e^{y}+c_{1}\end{align*}

but \(p=y^{\prime }\) hence the above becomes

\begin{align*} \left ( y^{\prime }\right ) ^{2} & =e^{y}+c_{1}\\ y^{\prime } & =\pm \sqrt {e^{y}+c_{1}}\end{align*}

This is quadrature. For the positive solution

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

Now we have to use (2) and take derivative and solve for \(c_{1},c_{2}\). Much harder than if we have applied IC to each solution earlier.