1. ## coupled differential equations

I really have no idea to go about doing question 2, I can't find any information about coupled differential equations that I understand, or any examples to help.

Here's question 1 to help with question 2

2. ## Re: coupled differential equations

In your first question, you have \displaystyle \begin{align*} \frac{dx}{dt} = y \end{align*}. Differentiating this w.r.t. t gives \displaystyle \begin{align*} \frac{d^2x}{dt^2} = \frac{dy}{dt} \end{align*}.

Your second equation is \displaystyle \begin{align*} \frac{dy}{dt} = - 5x - 2y \end{align*}. Substituting from the results from manipulating the first equation gives

\displaystyle \begin{align*} \frac{d^2x}{dt^2} &= -5x - 2\,\frac{dx}{dt} \\ \frac{d^2x}{dt^2} + 2\,\frac{dx}{dt} + 5x &= 0 \end{align*}

which is a second order linear constant coefficient ODE. Solve as normal. When you have x, you should be able to find y.

3. ## Re: coupled differential equations

However, because (1a) asks you to find eigenvalues and eigen vectors of the matrix $\begin{pmatrix}0 & 1 \\ -5 & 2 \end{pmatrix}$ and problem (2b) says "Hence write down the general solution of the second order differential equation" I suspect the student is not intended to change the system of equations to the second order equation but solve the corresponding matrix equation
$\begin{pmatrix}\frac{dx}{dt} \\ \frac{dy}{dt}\end{pmatrix}= \begin{pmatrix}0 & 1 \\ -5 & -2\end{pmatrix}$
or "dX/dt= AX"

In problem (1a) we are given that one eigenvalue is 1- 2i and can then find that the other is -1+ 2i. We can then find the corresponding eigenvectors. Letting P be the matrix having those eigenvectors as columns, we have " $A= P^{-1}DP$ where D is the diagonal matrix\[tex]\begin{pmatrix}1- 2i & 0 \\ 0 & -1+ 2i\end{pmatrix}
Then we can write the differential equation as dX/dt= AX= P^{-1}DPX or, multiplying on both sides by P, PdX/dt= d(PX)/dt= D(PX) which, letting Z= PX, is dZ/dt= DZ. That now is uncoupled:
$\frac{dZ}{dt}= \begin{pmatrix}\frac{dz_1}{dt} \\ \frac{dz_2}{dt}\end{pmatrix}= \begin{pmatrix} (1- 2i)z_1 \\ (-1+ 2i)z_2\end{pmatrix}$
$\frac{dz_1}{dt}= (1- 2i)z_1$ is easy to solve: $z_1(t)= C_1e^{(1- 2i)t= C_1e^t(cos(2t)- i sin(2t))$
$\frac{dz_2}{dt}= (-1+ 2i)z_2$ is easy to solve: $z_2(t)= C_2e^{(-1+ 2i)t}= C_2e^{-t}(cos(2t)+ i sin(2t))$

And then, since $Z= PX$, $X= P^{-1}Z$.