# Coupled non-homogenous eigenvalue problem help?

1. Apr 1, 2012

### de1337ed

1. The problem statement, all variables and given/known data

Equations:
$\frac{dv_{1}}{dt}$ = $-v_{1}$ - $\frac{2v_{2}}{3}$ + 1 + $\frac{t}{3}$
$\frac{dv_{2}}{dt}$ = $-2v_{2}$ - 1 - 2t

Initial Conditions:
$v_{1}(0)$ = 6
$v_{2}(0)$ = -6

2. The attempt at a solution

Defined the following:
v(t) =
[ $v_{1}(t)$
$v_{2}(t)$ ]

$\frac{dv(t)}{dt}$ =
[ $\frac{dv_{1}(t)}{dt}$
$\frac{dv_{2}(t)}{dt}$ ]

v(0)=
[ 6
-6 ]

s =
[ 1 + $\frac{t}{3}$
-1 - 2t ]

A =
[ -1 $\frac{-2}{3}$
0 -2 ]

Particular:

$\frac{dv(t)}{dt}$ = Av + s
Try v = $k_{0}$ + $k_{1}t$

And now I'm stuck. I know what to do in homogenous cases when we use $cxe^{\lambda t}$ because in that case, you can represent $\frac{dv(t)}{dt}$ as $\lambda v$
But what do I do in the non-homogeneous case when there is a polynomial driving function? Thank you.

2. Apr 2, 2012

### K^2

You do exactly the same thing as in homogeneous case. You want to write your equation as v'=Av+f(t), where v', v, and f(t) are vectors, and A is a 2x2 matrix. You've gotten that far. Now, you need to diagonalize A. So say you found matrix S such that Ad=S-1AS, where Ad is your diagonalized matrix.

Having that, multiply your equation by S-1 from the left on both sides. You get.

S-1v' = S-1Av + S-1 f(t)

Now I'm going to insert I=SS-1 between A and v.

S-1v' = S-1AS S-1v + S-1 f(t)

S-1v' = Ad S-1v + S-1 f(t)

Because Ad is diagonal, you can re-write the above as two completely independent driven first order linear ordinary differential equations. Their driving terms are completely separate, so you should have no trouble solving each independently. Then simply multiply your solution vector by S from the left, and you'll get your answer.