Example 4 y′′ = ∘ ____

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

Now we can use the IC on this ode, since the IC is on \(y^{\prime }\). Solving this as ﬁrst order gives

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

Applying IC, where \(p\left ( 0\right ) =y^{\prime }\left ( 0\right ) =1\) gives

\begin{align*} 1 & =\sinh \left ( c_{1}\right ) \\ c_{1} & =\operatorname {arcsinh}\left ( 1\right ) \end{align*}

Hence

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

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

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

Solving as ﬁrst order ode gives

\[ y\left ( x\right ) =\cosh \left ( x+\ln \left ( 1+\sqrt {2}\right ) \right ) +c_{2}\]