second order linear constant coeﬀ

2.2.6.1 second order linear constant coeﬀ

\[\begin {aligned} x^{\prime \prime }-4 x^{\prime }+5 x&=0 \end {aligned}\]

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 \(A=1, B=-4, C=5\). Let the solution be \(x=e^{\lambda t}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\lambda t}-4 \lambda \,{\mathrm e}^{\lambda t}+5 \,{\mathrm e}^{\lambda t} = 0 \tag {1} \]

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

\[ \lambda ^{2}-4 \lambda +5 = 0 \tag {2} \]

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 \(A=1, B=-4, C=5\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {4}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {-4^2 - (4) \left (1\right )\left (5\right )}\\ &= 2 \pm i \end{align*}

Hence

\begin{align*} \lambda _1 &= 2 + i\\ \lambda _2 &= 2 - i \end{align*}

Which simpliﬁes to

\begin{gather*} \begin {aligned} \lambda _1 &= 2+i\\ \lambda _2 &= 2-i \end {aligned} \end{gather*}

Since the roots are complex conjugate of each others, then let the roots be

\begin{align*} \lambda _{1,2} &= \alpha \pm i \beta \end{align*}

Where \(\alpha =2\) and \(\beta =1\). Therefore the ﬁnal solution using Euler relation can be written as

\begin{align*} x &= e^{\alpha t} \left ( c_1 \cos (\beta t) + c_2 \sin (\beta t) \right ) \end{align*}

Which becomes

\begin{align*} x &= e^{2 t}\left (c_1 \cos \left (t \right )+c_2 \sin \left (t \right )\right ) \end{align*}

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