2.2.7.2 second order linear exact ode

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

Entering second order linear exact ode solverAn ode of the form

\begin{align*} p \left (t \right ) x^{\prime \prime }+q \left (t \right ) x^{\prime }+r \left (t \right ) x&=s \left (t \right ) \end{align*}

is exact if

\begin{align*} p''(t) - q'(t) + r(t) &= 0 \tag {1} \end{align*}

For the given ode the above values are

\begin{align*} p(x) &= 1\\ q(x) &= 3\\ r(x) &= 0\\ s(x) &= 0 \end{align*}

Hence

\begin{align*} p''(x) &= 0\\ q'(x) &= 0 \end{align*}

Therefore (1) becomes

\begin{align*} 0- \left (0\right ) + \left (0\right )&=0 \end{align*}

This shows the ode is exact. Since the ode is exact, it can be written as

\begin{align*} \left (p \left (t \right ) x^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) x\right )' &= s(x) \end{align*}

Integrating the above gives

\begin{align*} p \left (t \right ) x^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) x&=\int {s \left (t \right )\, dt} \end{align*}

Substituting the values of \(p,q,r,s\) into the above results in

\begin{align*} x^{\prime }+3 x&=c_1 \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {1}{-3 x +c_1}d x &= dt\\ -\frac {\ln \left (-3 x +c_1 \right )}{3}&= t +c_2 \end{align*}

Solving for \(x\) from \(-\frac {\ln \left (-3 x+c_1 \right )}{3} = t +c_2\) gives

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

Phase plot