Second Order ODE with Complex’s Roots

Problem 123: Given the second order ODE

(1) $\begin{equation*} y'' + 12y' + 45 y = 0, \end{equation*}$

find the solutions $y_1(x)$ and $y_2(x)$ .

Solution: The general solution of a second order ODE with complex root is given by

(2) $\begin{equation*} y = y_1(x) + y_2(x) = c_1 \cdot e^{r_1 x} + c_2 \cdot e^{r_2 x},\end{equation*}$

where $c_1$ and $c_2$ are arbitrary constant, $r_1$ and $r_2$ are the complex roots of the characteristic polynomial. Finding the roots of the characteristic polynomial gives us

(3) $\begin{align*} y'' + 12y' + 45 y & = 0 \\ r^2 + 12r + 45 & = 0 \\ & \Rightarrow r = -6 \pm 3i. \quad \text{(the roots)}\end{align*}$

That is,

(4) $\begin{align*}y_1(x) & = c_1 \cdot e^{r_1 x} \\ & = c_1 \cdot e^{(-6+3i)x} \\ & = c_1 \cdot e^{-6x} \cdot e^{i(3x)} \\ & = c_1e^{-6x} \left( \cos(3x) + i \sin(3x)\right). \left \end{align*}$

Similarly,

(5) $\begin{align*}y_2(x) & = c_2 \cdot e^{r_2 x} \\ & = c_2 \cdot e^{(-6-3i)x} \\ & = c_2 \cdot e^{-6x} \cdot e^{i(-3x)} \\ & = c_2 e^{-6x}\left( \cos(-3x) + i \sin(-3x)\right) \\ & = c_2e^{-6x} \left( \cos(3x) - i \sin(3x)\right). \end{align*}$

NOTE: Since this is a linear ODE, one can add the two solutions in order to get a general solution, but I will just give the two solutions separately. Also, a little bit of more simplification can be obtained, but for simplicity it will not be done.

Hence,

(6) $\begin{align*} y_1(x) & = c_1e^{-6x} \left(\cos(3x) + i \sin(3x) \right, \quad \text{and} \\ y_2(x) & = c_2e^{-6x} \left(\cos(3x) - i \sin(3x) \right). \end{align*}$

One Problem at a Time

Second Order ODE with Complex’s Roots

Leave a Reply Cancel reply