Example 3A y′′ = ∘ ____

\begin{align} y^{\prime \prime } & =\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\tag {1}\\ y\left ( 0\right ) & =1\nonumber \end{align}

Notice that only one IC is given. Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\). Hence the ode becomes

\begin{equation} p^{\prime }=\sqrt {1+p^{2}} \tag {2}\end{equation}

We can’t use IC on this ode, since the IC is only on \(y\) and not \(y^{\prime }\). Solving this as ﬁrst order gives

\[ p\left ( x\right ) =\sinh \left ( x+c_{1}\right ) \]

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

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

Now we solve this using the IC \(y\left ( 0\right ) =1\). Solving the above gives

\begin{equation} y=\cosh \left ( x+c_{1}\right ) +c_{2} \tag {3}\end{equation}

Applying IC, and now we need to be careful. We need to solve for \(c_{2}\) and not \(c_{1}\).

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

Hence (3) becomes

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