2.2.7.3 second order ode missing y

\[\begin {aligned} x^{\prime \prime }+3 x^{\prime }&=0 \end {aligned}\]

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

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

Then

\begin{align*} u'(t) &= x^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} u^{\prime }\left (t \right )+3 u \left (t \right ) = 0 \end{align*}

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

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int -\frac {1}{3 u}d u &= dt\\ -\frac {\ln \left (u \right )}{3}&= t +c_1 \end{align*}

Solving for \(u \left (t \right )\) from \(-\frac {\ln \left (u \left (t \right )\right )}{3} = t +c_1\) gives

\begin{align*} u \left (t \right ) &= {\mathrm e}^{-3 t -3 c_1} \\ \end{align*}

For solution \(u \left (t \right ) = {\mathrm e}^{-3 t -3 c_1}\), since \(u=x^{\prime }\) then the new first order ode to solve is

\begin{align*} x^{\prime } = {\mathrm e}^{-3 t -3 c_1} \end{align*}

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

\begin{align*} dx &= \left ({\mathrm e}^{-3 t -3 c_1}\right ) \, dt\\ x &= \int { \left ({\mathrm e}^{-3 t -3 c_1}\right ) \, dt}\\ &= -\frac {{\mathrm e}^{-3 t -3 c_1}}{3}+c_2 \end{align*}

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

\begin{gather*} \begin {aligned} x &= -\frac {{\mathrm e}^{-3 t -3 c_1}}{3}+c_2\\ \end {aligned} \end{gather*}

Phase plot