2.2.5.1 second order linear constant coeff

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

Entering second order linear constant coefficient ode solverThis is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A x''(t) + B x'(t) + C x(t) = 0 \]

Where in the above \(A=1, B=-4, C=4\). Let the solution be \(x=e^{\lambda t}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\lambda t}-4 \lambda \,{\mathrm e}^{\lambda t}+4 \,{\mathrm e}^{\lambda t} = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda t}\) gives

\[ \lambda ^{2}-4 \lambda +4 = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form. Using the quadratic formula the roots are

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=-4, C=4\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {4}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {\left (-4\right )^2 - (4) \left (1\right )\left (4\right )}\\ &= 2 \end{align*}

Hence this is the case of a double root \(\lambda _{1,2} = -2\). Therefore the solution is

\[ x= c_1 {\mathrm e}^{2 t} + c_2 t \,{\mathrm e}^{2 t} \tag {1} \]

Phase plot