# Thread: The Method of Elimination

1. ## The Method of Elimination

FIND THE GENERAL SOLUTION OF THE LINEAR SYSTEM

x'' = 6x + 2y, y'' = 3x + 7y

2. Originally Posted by bearej50
FIND THE GENERAL SOLUTION OF THE LINEAR SYSTEM

x'' = 6x + 2y, y'' = 3x + 7y
Define u= x' and v= y'. Then the differential equation x"= 6x+ 2y becomes u'= 6x+ 2y and y"= 3x+ 7y becomes v'= 3x+ 7y.

So instead of two second order equations we have four first order equations:
x'= u, y'= v, u'= 6x+ 2y, and v'= 3x+ 7y which we could also write as the single first order matrix equation:
$\displaystyle \frac{d\begin{bmatrix}x \\ y \\ u \\ v\end{bmatrix}}{dt}= \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 1 \\ 6 & 2 & 0 & 0 \\3 & 7 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ u \\ v\end{bmatrix}$

And the first step in that is to find the eigenvalues and eigenvectors of the coefficient matrix.

3. Originally Posted by bearej50
FIND THE GENERAL SOLUTION OF THE LINEAR SYSTEM

x'' = 6x + 2y, y'' = 3x + 7y
As the title of the post suggests, if we let $\displaystyle y = \frac{x''-6x}{2}$ (**) and eliminate y from the second ODE we obtain $\displaystyle x^{(4)} - 12x'' + 36 = 0$. Its characteristic equation is $\displaystyle m^4 - 13m^2 + 36 = 0$ or $\displaystyle (m^2-4)(m^2-9)=0$ which has solution $\displaystyle m = \pm2, \pm3$ and so the solution is

$\displaystyle x = c_1 e^{-2x} + c_2 e^{2x} + c_3 e^{-3x} + c_4 e^{3x}$

Once you have this, sub into your y above (**).