Decouple a system of two second order differential equations I have a system of second-order differential equations that I want to decouple. they are,
$\ddot{x} = \frac{\omega_1^2}{2} x + \omega_2 \dot{y}$
and
$\ddot{y} = \frac{\omega_1^2}{2} y - \omega_2 \dot{x}$
I am thinking that I should use some transformation, but it just isn't clear in my head yet!
Thanks!
 A: Differentiate the first wrt $t$ to gain an expression for $\ddot y$:
$\dddot{x} = \frac{\omega_1^2}{2} \dot x + \omega_2 \ddot{y}$
Substitute $\ddot{y} = \frac{\omega_1^2}{2} y - \omega_2 \dot{x}$ to get:
$\dddot{x} = \frac{\omega_1^2}{2} \dot x + \frac{\omega_1^2 \omega_2}{2} y - \omega_2^2 \dot{x}$
Rearrange: $\frac{\omega_1^2 \omega_2}{2} y =\dddot{x} - \frac{\omega_1^2}{2} \dot x + \omega_2^2 \dot{x}$
Differentiate: $\frac{\omega_1^2 \omega_2}{2} \dot y =\ddddot{x} + \frac{2\omega_2^2 -\omega_1^2}{2} \ddot x$
Recall that $\ddot{x} = \frac{\omega_1^2}{2} x + \omega_2 \dot{y} \Rightarrow  \omega_2 \dot{y}=\ddot{x} - \frac{\omega_1^2}{2} x$
Thus: $\frac{\omega_1^2}{2} \left (\ddot{x} - \frac{\omega_1^2}{2} x \right ) =\ddddot{x} + \frac{2\omega_2^2 -\omega_1^2}{2} \ddot x$
... which becomes $\ddddot{x} + \frac{2\omega_2^2 -\omega_1^2}{2} \ddot x -  \frac{\omega_1^2}{2} \left (\ddot{x} - \frac{\omega_1^2}{2} x \right )=0 $
or $\ddddot{x} + \left(\omega_2^2 -\omega_1^2 \right ) \ddot x -  \frac{\omega_1^4}{2}x =0 $
Auxiliary equation $\lambda^4+p\lambda^2-q=0$ where $p=\left(\omega_2^2 -\omega_1^2 \right )$ and $q=\omega_1^4$
$\lambda^2={-p+\sqrt{p^2+4q} \over 2}$ or $\lambda^2={-p-\sqrt{p^2+4q} \over 2}$
$\lambda_1=\sqrt{{-p+\sqrt{p^2+4q} \over 2}}$
$\lambda_2=-\sqrt{{-p+\sqrt{p^2+4q} \over 2}}$
$\lambda_3=\sqrt{{-p-\sqrt{p^2+4q} \over 2}}$
$\lambda_4=-\sqrt{{-p-\sqrt{p^2+4q} \over 2}}$
A: Using Laplace transform:


*

*$$x''(t)=\frac{\omega_1^2}{2}\cdot x(t)+\omega_2y'(t)\Longleftrightarrow s^2x(s)-sx(0)-x'(0)=\frac{\omega_1^2}{2}\cdot x(s)+\omega_2\left[sy(s)-y(0)\right]$$

*$$y''(t)=\frac{\omega_1^2}{2}\cdot y(t)-\omega_2x'(t)\Longleftrightarrow s^2y(s)-sy(0)-y'(0)=\frac{\omega_1^2}{2}\cdot y(s)-\omega_2\left[sx(s)-x(0)\right]$$


Now, we can find that:


*

*$$x(s)=\frac{2\left(sx(0)+x'(0)-\omega_2y(0)+\omega_2y(s)s\right)}{2s^2-\omega_1^2}$$

*$$y(s)=\frac{2\left(sy(0)+y'(0)-\omega_2x(0)+\omega_2x(s)s\right)}{2s^2-\omega_1^2}$$


Substitute them into each other and solve it with inverse Laplace transform.
