The Method of Elimination

**bearej50** · May 5th 2009, 04:10 PM

FIND THE GENERAL SOLUTION OF THE LINEAR SYSTEM

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

**HallsofIvy** · May 5th 2009, 04:45 PM

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:
$\frac{d\begin{bmatrix}x \\ y \\ u \\ v\end{bmatrix}}{dt}= \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 1 \\ 6 & 2 & 0 & 0 \\3 & 7 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ u \\ v\end{bmatrix}$

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

**Danny** · May 6th 2009, 06:24 AM

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

$<br /> x = c_1 e^{-2x} + c_2 e^{2x} + c_3 e^{-3x} + c_4 e^{3x}<br />$

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

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