Solving a coupled pair of differential equations using matrices

I am asked to find the eigenvalues and eigenvectors of the matrix: $\displaystyle A = \left(\begin{array}{cc}5&3\\1&7\end{array}\right)$

I find them as $\displaystyle \lambda_1 = 8 $, $\displaystyle \lambda_2 = 4 $ and $\displaystyle v_1 = (1,1) $, $\displaystyle v_2 = (-3,1) $

The equations I need to solve are: $\displaystyle \frac{dx}{dt} = 5x +3 y $ and $\displaystyle \frac{dy}{dt} = x + 7y$. These equations are coupled (whatever that means...)

I turn these into a pair of simultaneous equations sort of thing to form the matrix equation $\displaystyle X' = AX $, where X' is $\displaystyle \left(\begin{array}{cc}\frac{dx}{dt}\\\frac{dy}{dt }\end{array}\right)$ and $\displaystyle X = \left(\begin{array}{cc}x&\\y&\end{array}\right)$

As far as I know, I have diagonalise A which will give me $\displaystyle A = P^{-1}DP$. I know $\displaystyle P = \left(\begin{array}{cc}1&-3\\1&1\end{array}\right)$ and $\displaystyle D = \left(\begin{array}{cc}8&0\\0&4\end{array}\right)$ using the eigenvalues.

So I go along...

$\displaystyle X' = P^{-1}DPX $

$\displaystyle PX' = DPX $

$\displaystyle (PX)' = D(PX) $

Subbing matrices in...

$\displaystyle \left(\begin{array}{cc}1&-3\\1&1\end{array}\right) * \left(\begin{array}{cc}\frac{dx}{dt}\\\frac{dy}{dt }\end{array}\right) = \left(\begin{array}{cc}8&0\\0&4\end{array}\right)* \left(\begin{array}{cc}1&-3\\1&1\end{array}\right)*\left(\begin{array}{cc}x& \\y&\end{array}\right)$

Can someone check if what I've done is correct so far? And if so, am I allowed to expend all the differentials and all the matrices then treat them as simultaneous equations. Also, I'm not too sure how to get rid of the $\displaystyle dt$ when there is no $\displaystyle t$ variable in the equations...

Re: Solving a coupled pair of differential equations using matrices

(The two equations are "coupled" means that each equation involve both of the functions.)

Yes, what you have done so far is correct.

The point is that letting Z= PX, you have

$\displaystyle \frac{dZ}{dt}= DZ$

$\displaystyle \begin{pmatrix}\frac{dz_1}{dt} \\ \frac{dz_2}{dt}\end{pmatrix}= \begin{pmatrix}8 & 0 \\ 0 & 4 \end{pmatrix}\begin{pmatrix{z_1 \\ z_2 \end{pmatrix}$

$\displaystyle \begin{pmatrix}\frac{dz_1}{dt} \\ \frac{dz_2}{dt}\end{pmatrix}= \begin{pmatrix}8z_1 \\ 4z_2\end{pmatrix}$

which is equivalent to the **two** "uncoupled" equations

$\displaystyle \frac{dz_1}{dt}= 8z_1$ and $\displaystyle \frac{dz_2}{dt}= 4z_2$

Those obviously have solutions $\displaystyle z_1= C_1e^{8t}$ and $\displaystyle z_2= C_2e^{4t}$ so that

$\displaystyle Z(t)= \begin{pmatrix}C_1e^{8t} \\ C_2e^{4t}\end{pmatrix}$

Since $\displaystyle Z= PX$, $\displaystyle X= P^{-1}Z$:

$\displaystyle \begin{pmatrix}x(t) \\ y(t)\end{pmatrix}= \begin{pmatrix}\frac{1}{4} & \frac{3}{4} \\ \frac{-1}{4} & \frac{1}{4} \end{pmatrix}\begin{pmatrix}C_1e^{8t} \\ C_2e^{4t}\end{pmatrix}$

$\displaystyle \begin{pmatrix}x(t) \\ y(t) \end{pmatrix}= \begin{pmatrix}\frac{1}{4}C_1e^{8t}+ \frac{3}{4}C_2e^{4t} \\ -\frac{1}{4}C_1e^{8t}+ \frac{1}{4}C_2e^{4t}\end{pmatrix}$

Re: Solving a coupled pair of differential equations using matrices

I hope I don't break the rules by posting on this but I believe this question is valid.

I know that it can be proven that the general solution to the D.E can be written as $\displaystyle x(t) = C_{1}e^{\lambda_{1}*t}v_{1} + C_{2}e^{\lambda_{2}*t}v_2$. I just learnt this but I'm not too sure where it came from.

So that makes $\displaystyle x(t) = C_{1}e^{8x} - 3C_{2}e^{4x} $ and $\displaystyle y(x) = C_{1}e^{8x} + C_{2}e^{4x} $ which is different to the solutions given by diagonalising matrix A. Am I misunderstanding something here?