2.6.3.1 second order ode missing x

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

Entering second order ode missing \(x\) solverThis is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = \frac {m \sqrt {1+p \left (y \right )^{2}}}{k} \end{align*}

Or

\begin{align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = \frac {m \sqrt {1+p \left (y \right )^{2}}}{k} \end{align*}

Which is now solved as first order ode for \(p(y)\).

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {k p}{m \sqrt {p^{2}+1}}d p &= dy\\ \frac {\sqrt {p^{2}+1}\, k}{m}&= y +c_1 \end{align*}

Solving for \(p\) from \(\frac {\sqrt {1+p^{2}}\, k}{m} = y +c_1\) gives

\begin{align*} p &= \frac {\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y +m^{2} y^{2}-k^{2}}}{k} \\ p &= -\frac {\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y +m^{2} y^{2}-k^{2}}}{k} \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = \frac {\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y+m^{2} y^{2}-k^{2}}}{k} \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {k}{\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y +m^{2} y^{2}-k^{2}}}d y &= dx\\ \frac {k \ln \left (\frac {c_1 \,m^{2}+m^{2} y}{\sqrt {m^{2}}}+\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y +m^{2} y^{2}-k^{2}}\right )}{\sqrt {m^{2}}}&= x +c_2 \end{align*}

Solving for \(y\) from \(\frac {k \ln \left (\frac {c_1 \,m^{2}+y m^{2}}{\sqrt {m^{2}}}+\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y+m^{2} y^{2}-k^{2}}\right )}{\sqrt {m^{2}}} = x +c_2\) gives

\begin{align*} y &= \frac {\left (-2 c_1 \,m^{2} {\mathrm e}^{\frac {\sqrt {m^{2}}\, \left (x +c_2 \right )}{k}}+\sqrt {m^{2}}\, k^{2}+{\mathrm e}^{\frac {2 \sqrt {m^{2}}\, \left (x +c_2 \right )}{k}} \sqrt {m^{2}}\right ) {\mathrm e}^{-\frac {\sqrt {m^{2}}\, \left (x +c_2 \right )}{k}}}{2 m^{2}} \\ \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = -\frac {\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y+m^{2} y^{2}-k^{2}}}{k} \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int -\frac {k}{\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y +m^{2} y^{2}-k^{2}}}d y &= dx\\ -\frac {k \ln \left (\frac {c_1 \,m^{2}+m^{2} y}{\sqrt {m^{2}}}+\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y +m^{2} y^{2}-k^{2}}\right )}{\sqrt {m^{2}}}&= x +c_3 \end{align*}

Solving for \(y\) from \(-\frac {k \ln \left (\frac {c_1 \,m^{2}+y m^{2}}{\sqrt {m^{2}}}+\sqrt {c_1^{2} m^{2}+2 c_1 \,m^{2} y+m^{2} y^{2}-k^{2}}\right )}{\sqrt {m^{2}}} = x +c_3\) gives

\begin{align*} y &= \frac {\left (-2 c_1 \,m^{2} {\mathrm e}^{-\frac {\sqrt {m^{2}}\, \left (x +c_3 \right )}{k}}+\sqrt {m^{2}}\, k^{2}+{\mathrm e}^{-\frac {2 \sqrt {m^{2}}\, \left (x +c_3 \right )}{k}} \sqrt {m^{2}}\right ) {\mathrm e}^{\frac {\sqrt {m^{2}}\, \left (x +c_3 \right )}{k}}}{2 m^{2}} \\ \end{align*}