4.9.1.1.13 Example 13 \(y^{\prime \prime }=-e^{-2y}=0,y\left ( 3\right ) =0,y^{\prime }\left ( 3\right ) =1\)

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

With IC

\begin{align*} y\left ( 3\right ) & =0\\ y^{\prime }\left ( 3\right ) & =1 \end{align*}

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes

\begin{equation} p\frac {dp}{dy}=-e^{-2y}\nonumber \end{equation}

This is separable.

\begin{align} \int pdp & =-\int e^{-2y}dy\nonumber \\ \frac {p^{2}}{2} & =\frac {1}{2}e^{-2y}+c_{1}\nonumber \\ p^{2} & =e^{-2y}+c_{2}\tag {3}\end{align}

Which gives

\begin{align*} p_{1} & =\sqrt {e^{-2y}+c_{2}}\\ p_{2} & =-\sqrt {e^{-2y}+c_{2}}\end{align*}

Before solving this, we should apply IC now as it simplifies the solution greatly. This assumes both \(y,y^{\prime }\) are are given at same point \(x_{0}\). Which is the case here. If only one IC is given (such as \(y\left ( 0\right ) \) or \(y^{\prime }\left ( 0\right ) \) but not both, then we can not apply IC now and have to do it at the end).

We are given that \(y^{\prime }\left ( 3\right ) =p=1,y\left ( 3\right ) =0\), hence the IC becomes \(p\left ( 0\right ) =1\). Above gives

\begin{align*} p_{1} & =\sqrt {e^{-2y}+c_{2}}\\ 1 & =\sqrt {1+c_{2}}\\ c_{2} & =0 \end{align*}

And for the second one

\begin{align*} p_{2} & =-\sqrt {e^{-2y}+c_{2}}\\ 1 & =-\sqrt {1+c_{2}}\end{align*}

No solution. Therefore we have

\begin{align*} p_{1} & =\sqrt {e^{-2y}}\\ & =e^{-y}\end{align*}

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

\[ y^{\prime }=e^{-y}\]

This is quadrature. Solving gives

\[ y=\ln \left ( x+c_{1}\right ) \]

Applying IC For \(y\left ( 3\right ) =0\) gives

\begin{align*} 0 & =\ln \left ( 3+c_{1}\right ) \\ 3+c_{1} & =1\\ c_{1} & =-2 \end{align*}

Hence first solution is

\[ y=\ln \left ( x-2\right ) \]

Similar steps for the negative root. This problem will become very hard to solve for the IC if we had not used the IC during solving the first ode.