Solving Systems of Linear Differential Equations

[shelovesmath]

1. x' + y' - x = -2t
x' + y' - 3x -y = t^2




2. x' = Dx, y' = Dy




3.
I eliminated x to get y alone by multiplying the first row by (D-3) and the second row by (D-1)

(D-1)(D-3)x + (D-3)Dy = -2t(D-3)
(D-1)(D-3)x + (D-1)(D-1) = t^2(D-1)

then subtracted to get: (D+1)y = -t^2 -4t + 2

I put this into a differential equation form y' + y = -t^2 - 4t + 2 This is linear first order, and my integrating factor is e^t (at this point I will solve this linear ODE)

For getting x by itself, I eliminated y by multiplying the first row by (D-1) and the second row by D

(D-1)(D-1)x + D(D-1)y = -2t(D-1)
D(D-3)x + D(D-1)y = Dt^2

then subtracted to get: (D+1)y = -2

I put this into a differential equation form y' + y = -2 This is linear first order, and my integrating factor is also e^t (at this point I will solve this linear ODE)



I just need to know that I'm good up to here.

 

[grey_earl]

First: your notation is a bit confusing. Also derivative operators (usually) act on the right, so you multiply them from the left, and therefore have -2 (D-3) t instead of -2 t (D-3). However, your result is correct, that is
y' + y = -t^2 - 4t + 2. (the integrating factor e^t is also correct)

Afterwards (in the second calculation) I think you confused y and x, since I get
x' + x = -2.

 

[Antiphon]

I recommend writing it in matrix notation.

 

[shelovesmath]

I recommend writing it in matrix notation.

I have to use this method specifically for my homework assignment.