# Thread: Solution for system of diff equations

1. ## Solution for system of diff equations

Find the solution of the system
x' = 3x + 4y; y' = -2x-3y
satisfying x(0) = 2 and y(0) = -1
I turned 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? ..adn yes I have to use eigenvectors and eigenvalues

2. Originally Posted by PatrickM
Find the solution of the system
x' = 3x + 4y; y' = -2x-3y
satisfying x(0) = 2 and y(0) = -1
I turned 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? ..adn yes I have to use eigenvectors and eigenvalues
$\displaystyle \begin{pmatrix} x \\ y \end{pmatrix}_{t} = \begin{pmatrix}3 & 4 \\ -2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

Now to find the eigenvalues

$\displaystyle \begin{vmatrix}3-\lambda & 4 \\ -2 & -3-\lambda \end{vmatrix}=(3-\lambda)(-3-\lambda)+8=\lambda^2 -9+8=(\lambda+1)(\lambda-1)$

Now to find the eigenvectors

when $\displaystyle \lambda =1$

$\displaystyle \begin{pmatrix}2 & 4 \\ -2 & -4 \end{pmatrix} \iff \begin{pmatrix}1 & 2 \\ 0 & 0 \end{pmatrix}$

Now set $\displaystyle y=t \implies x=-2y=-2t$

$\displaystyle \begin{pmatrix}x \\ y \end{pmatrix} =\begin{pmatrix}-2t \\ t \end{pmatrix}=\begin{pmatrix}-2 \\ 1 \end{pmatrix}t$

when $\displaystyle \lambda =-1$

$\displaystyle \begin{pmatrix}4 & 4 \\ -2 & -2 \end{pmatrix} \iff \begin{pmatrix}1 & 1 \\ 0 & 0 \end{pmatrix}$

Now set $\displaystyle y=t \implies x=-y=-t$

$\displaystyle \begin{pmatrix}x \\ y \end{pmatrix} =\begin{pmatrix}-t \\ t \end{pmatrix}=\begin{pmatrix}-1 \\ 1 \end{pmatrix}t$

So the general solution is

$\displaystyle \begin{pmatrix}x(t) \\ y(t) \end{pmatrix}=c_1\begin{pmatrix}-2 \\ 1 \end{pmatrix}e^{t}+c_2\begin{pmatrix}-1 \\ 1 \end{pmatrix}e^{-t}$

3. thank you TheEmptySet