# Real Solution of 2nd Order ODE (Complex Roots)

• Aug 8th 2009, 05:52 PM
Maccaman
Real Solution of 2nd Order ODE (Complex Roots)
I've put the ODE into a system.

We have:
$\left[\begin{array}{cc}\dot{y_1}\\\dot{y_2}\end{array}\r ight] = \left[\begin{array}{cc}-1&-4\\1&3\end{array}\right] \left[\begin{array}{cc}y_1\\y_2\end{array}\right]$

which has Eigenvalues:
$\lambda_1 = -2 + i \sqrt{3}$
$\lambda_2 = -2 - i \sqrt{3}$

and Eigenvectors:
$x^{(1)} = \left[\begin{array}{cc}1+i \sqrt{3}\\1\end{array}\right]$, $x^{(2)} = \left[\begin{array}{cc}1-i \sqrt{3}\\1\end{array}\right]$

General Solution is
$y(t) = c_1 \left[\begin{array}{cc}1+i \sqrt{3}\\1\end{array}\right] e^{(-2+i \sqrt{3})t} + c_2 \left[\begin{array}{cc}1-i \sqrt{3}\\1\end{array}\right]e^{(-2-i \sqrt{3})t}$

But this is where I am getting lost, I know that when the eigenvalues are complex I need to involve sin and cos, which is fine for a 1st order ODE but am getting a little stuck with this 2nd-order ODE.

Also, I need to find the solution in Real Form. i.e.

$y(t) = \left[\begin{array}{cc}y_1(t)\\y_2(t)\end{array}\right]$
Thanks in advance to anyone who can help.
• Aug 9th 2009, 12:39 AM
Rebesques
The real and imaginary parts of the general solution are two (linearly independent) real solutions.
• Aug 12th 2009, 09:37 PM
Maccaman
I was wondering if someone could check my solution. If its not correct, I think it could be close.....

General Solution:
$
y(t) = c_1 \begin{pmatrix}1+i \sqrt{3}\\1\end{pmatrix} e^{(-2+i \sqrt{3})t} + c_2 \begin{pmatrix}1-i \sqrt{3}\\1\end{pmatrix}e^{(-2-i \sqrt{3})t}
$

$
= e^{-2t} \Bigg ( c_1 \begin{pmatrix}1+i \sqrt{3}\\1\end{pmatrix} e^{(\sqrt{3})it} + c_2 \begin{pmatrix}1-i \sqrt{3}\\1\end{pmatrix} e^{(-\sqrt{3})it} \Bigg )
$

$
= e^{-2t} \Bigg ( c_1 \begin{pmatrix}1+i \sqrt{3}\\1\end{pmatrix}(c_1 cos(\sqrt{3}t)+i sin(\sqrt{3}t)+c_2 \begin{pmatrix}1-i \sqrt{3}\\1\end{pmatrix}(cos(\sqrt{3}t)-i sin(\sqrt{3}t) \Bigg)
$
(Complex Form)

Note: Don't confuse $cos(\sqrt{3}t)$ and $sin(\sqrt{3}t)$ with $cos(\sqrt{3t})$ and $sin(\sqrt{3t})$

$= e^{-2t} \begin{pmatrix} a \\b+c \end{pmatrix}$

where $a = c_1 cos(\sqrt{3} t) + i c_1 sin (\sqrt{3} t) + c_2 cos(\sqrt{3} t) - i c_2 sin (\sqrt{3} t)$

and $b = (\sqrt{3})i \ c_1 cos(\sqrt{3} t) - c_1 cos(\sqrt{3} t) - (\sqrt{3}) c_1 sin(\sqrt{3} t) - i c_1 sin(\sqrt{3} t)$

and $c = - (\sqrt{3}) i c_2 cos(\sqrt{3} t) - c_2 sin(\sqrt{3} t) - (\sqrt{3}) c_2 sin(\sqrt{3} t) + i c_2 sin(\sqrt{3} t)$

(I'm sorry if this part is confusing, but when I tried to write one piece of latex code for this I got an error message saying something like 'Latex Error too big')

$= (\sqrt{3}+c_2)e^{-2t} \begin{pmatrix} cos(\sqrt{3} t) \\ -cos(\sqrt{3}t)-(\sqrt{3})sin(\sqrt{3}t) \end{pmatrix}$ ....

..... $+ i(c_1-c_2)e^{-2t} \begin{pmatrix} sin((\sqrt{3}t) \\ (\sqrt{3})cos(\sqrt{3}t)-sin(\sqrt{3}t) \end{pmatrix}
$

$= c_1 e^{-2t} \begin{pmatrix}cos(\sqrt{3} t) \\ -cos(\sqrt{3}t)-(\sqrt{3})sin(\sqrt{3}t) \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix} sin((\sqrt{3}t) \\ (\sqrt{3})cos(\sqrt{3}t)-sin(\sqrt{3}t) \end{pmatrix}
$
(Real Form)

whew....(wipes brow)