- #1

shelovesmath

- 60

- 0

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

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

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

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

2. x' = Dx, y' = Dy

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)

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.