2.6.3.2 second order ode missing y

\[\begin {aligned} y^{\prime \prime }&=\frac {m \sqrt {1+{y^{\prime }}^{2}}}{k} \end {aligned}\]

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(x) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} u^{\prime }\left (x \right )-\frac {m \sqrt {1+u \left (x \right )^{2}}}{k} = 0 \end{align*}

Which is now solved for \(u(x)\) as first order ode.

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {k}{m \sqrt {u^{2}+1}}d u &= dx\\ \frac {k \,\operatorname {arcsinh}\left (u \right )}{m}&= x +c_1 \end{align*}

Solving for \(u \left (x \right )\) from \(\frac {k \,\operatorname {arcsinh}\left (u \left (x \right )\right )}{m} = x +c_1\) gives

\begin{align*} u \left (x \right ) &= \sinh \left (\frac {m \left (x +c_1 \right )}{k}\right ) \\ \end{align*}

For solution \(u \left (x \right ) = \sinh \left (\frac {m \left (x +c_1 \right )}{k}\right )\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = \sinh \left (\frac {m \left (x +c_1 \right )}{k}\right ) \end{align*}

Entering first order ode quadrature solverBecause the ODE has the form \(y^{\prime }=f(x)\), the solution requires only integration. Therefore

\begin{align*} dy &= \left (\sinh \left (\frac {m \left (x +c_1 \right )}{k}\right )\right ) \, dx\\ y &= \int { \left (\sinh \left (\frac {m \left (x +c_1 \right )}{k}\right )\right ) \, dx}\\ &= \frac {k \cosh \left (\frac {m x}{k}+\frac {m c_1}{k}\right )}{m}+c_2 \end{align*}

In summary, these are the solution found for \(y\)

\begin{gather*} \begin {aligned} y &= \frac {k \cosh \left (\frac {m x}{k}+\frac {m c_1}{k}\right )}{m}+c_2\\ \end {aligned} \end{gather*}