Example 4 \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\[ y\left ( x\right ) =\cosh \left ( x+\ln \left ( 1+\sqrt {2}\right ) \right ) +c_{2}\]