First Order Linear Systems with constant coefficients

[forza1nter]

in solving this style of equation

x'1 = x1 + 2x2, x'2 = 4x1 -x2, x1(0)=5,x2=(0)= -2

what is the best approach to this problem and how do i set it up? what clues should i look for in these problems (if any)?
thx!!!!!!!!!

I/m

 

[Unco]

I presume you're not familiar with the eigenvalue-eigenvector method, so instead we'll consider turning the system into a second order linear homogeneous DE. By solving the first equation for x2, and differentiating to get x2', we have x2 and x2' each in terms of x1, x1' and x1''. Substituting these expressions for x2 and x2' into the second equation (of your system) gives a second order DE in x1 only, which you can then solve for x1. Finally, you can use that to solve for x2.

Don't forget to show your complete work if you get stuck.

 

[forza1nter]

ok, this is what ive done so far....think im close, but still stuck (or confused)...sure i made a small error so any help is needed:

X’1 = x1 + 2x2
X’2 = 4x1 - x2

where x1 (0) = 5 and x2 (0) = -2

use operator form:
X'1 - X1 -2X = 0
-4X + X'2 + X2 = 0
(D -1) X1 - 2X2 = 0
-4X1 + (D+1)X2 = 0
-4X1 + (D+1)(D+1)X1 = 0
-4X1 + (D^2 - 1)X1 = 0.......im lost

I think i got something wrong, but having difficult time recognizing the mistake, please help w/ solution b/c now i'm worried...thx!!!!!!!!!!

I/m

 

[Unco]

ok, this is what ive done so far....think im close, but still stuck (or confused)...sure i made a small error so any help is needed:

X’1 = x1 + 2x2 (*)
X’2 = 4x1 - x2

where x1 (0) = 5 and x2 (0) = -2

use operator form:
X'1 - X1 -2X = 0
-4X + X'2 + X2 = 0
(D -1) X1 - 2X2 = 0
-4X1 + (D+1)X2 = 0 (#)

From (*), you can substitute x_2 = 1/2*x_1' - 1/2*x_1 into (#).

 

[forza1nter]

X’_1 = x_1 + 2x_2 (*)
X’_2 = 4x_1 - x_2

where x1 (0) = 5 and x2 (0) = -2

use operator form:
X'_1 - X_1 -2_X = 0
-4_X + X'_2 + X_2 = 0
(D -1) X_1 - 2X_2 = 0
-4X_1 + (D+1)X_2 = 0 (#)
now....
-4X_1 + (D+1)((D-1)X_1)/2 = 0
-4X_1 + 1/2(D^2 - 1)X_1 =0
(D^2 - 4)X_1 =0 --->X_1'' - 4X_1=0

CHAR EQN: r^2-4=0
r=2; r=-2

X_1 = C_1(e^2t) + C_2(te^-2t)

after that, using the product rule for x_2...it blows up....stuck again

 

[Unco]

-4X_1 + 1/2(D^2 - 1)X_1 =0
(D^2 - 4)X_1 =0

This step isn't right. You should get x_1'' - 9x_1 = 0.