2.2.5.2 second order ode solved by an integrating factor

\[\begin {aligned} x^{\prime \prime }-4 x^{\prime }+4 x&=0 \end {aligned}\]

Entering second order ode solved by an integrating factor solverThe ode satisfies this form

\[ x^{\prime \prime }+p \left (t \right ) x^{\prime }+\frac {\left (p \left (t \right )^{2}+p^{\prime }\left (t \right )\right ) x}{2} = f \left (t \right ) \]

Where \( p(t) = -4\). Therefore, there is an integrating factor given by

\begin{align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int -4 \, dx} \\ &= {\mathrm e}^{-2 t} \end{align*}

Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of the ODE a complete differential

\begin{gather*} \begin {aligned} \left ( M(x) x \right )'' &= 0\\ \left ( {\mathrm e}^{-2 t} x \right )'' &= 0 \end {aligned} \end{gather*}

Integrating once gives

\[ \left ( {\mathrm e}^{-2 t} x \right )' = c_1 \]

Integrating again gives

\[ \left ( {\mathrm e}^{-2 t} x \right ) = c_1 t +c_2 \]

Hence the solution is

\begin{align*} x &= \frac {c_1 t +c_2}{{\mathrm e}^{-2 t}} \end{align*}

Or

\begin{align*} x &= c_1 t \,{\mathrm e}^{2 t}+c_2 \,{\mathrm e}^{2 t} \end{align*}

Phase plot