# Homogeneous linear systems with constant coefficients.

## Homework Statement

Find the solution of the system
x' = 3x + 4y; y' = -2x-3y
satisfying x(0) = 2 and y(0) = -1
Thank you

## The Attempt at a Solution

My method involves linear algebra:
turn the pair of equations into a single first order vector differential equation of the form
v' = Av
where v is a column vector with entries x and y and A is a 2x2 matrix.
A= 3 4
-2 -3
=>det(A-Ir),where r is a constant and I is the identety matrix....=(3-r)(-3-r)+8=r^2-1=>
r1=-1 and r2=1 these are the eigenvectors.Was my approach ok?
Finding egenvectors:
(A-Ir)w_j=|3-1 4 | w1 = 0
|-2 -3+1 | w2 0
this result the system 2w_1+4w_2=0 and -2w_1-2w_2=0 where both w's =0
so the general solution is u=a_1 e^t *|0| +a_2e^-t|0|
|0| |0|
I know i did a mistake here ,was r1=i and r2=-i?

## Answers and Replies

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dextercioby
Science Advisor
Homework Helper
Is it mandatory to use a certain method to solve the system ? Try reducing your system to a single ODE, either for x, or for y. The resulting equation (also with constant coefficents) will be second order and can be easily integrated.

I have to use this method,yes

HallsofIvy
Science Advisor
Homework Helper
So your differential equation is
$$\frac{d\begin{pmatrix}x \\ y\end{pmatrix}}{dt}= \begin{pmatrix}3 & 4 \\ -2 & -3}\begin{pmatrix}x \\ y\end{pmatrix}$$

Yes, the eigenvalues are 1 and -1. However, you have found the eigenvectors wrong- the eigenvectors are vectors giveing [math]Av= \lambda v[/itex]. For $\lambda= 1$, that is
$$\begin{pmatrix}3 & 4 \\ -2 & -3\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}x \\ y \end{pmatrix}$$

$$\begin{pmatrix}2- 1 & 4 \\ -2 & -3- 1\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}2 & 4 \\ -2 & -4\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}0 \\ 0\end{pmatrix}$$
which is the same as 2x+ 4y= 0, -2x- 4y= 0 which have any x, y such that x= -2y as solution.

Do see what you did wrong? You subtracted 1 from one diagonal number and -1 from the other. Instead subtract 1 from both diagonal numbers to get an eigenvector corresponding to eigenvalue 1 and subtract -1 from both diagonal numbers to get an eigenvector corresponding to eigenvalue -1.

(The definition of "eigenvalue" is that there exist a non-zero vector, v such that $Av= \lambda v$. If you get 0 as the only solution, you have done something wrong!)