Verifying Laplace Transform Solution for x' + 2y' + x = 0, x' - y' + y = 1

[highlander2k5]

Can someone tell me if I did this right because my solution seems wrong, but I've done it a couple times and get the same answer. I'm given the following:
x' + 2y' + x = 0
x' - y' + y = 1
and the initial values of x(0) = 0 and y(0) = 1
The idea is to solve this initial value problem.

Here's my work.
Start by taking laplace transforms, so:
sX + 2sY - 2 + X = 0
sX - sY + 1 + Y = 1/s

D = | s+1 2s | = -3s^2 +1
| s -s+1 |

D_x = | 2 2s | = 0
| 1/s - 1 -s+1 |

D_y= | s+1 2 | = -3s^2 +1 / s
| s 1-s/s |

In between the | | are values to take cross product.

Then use Cramer's Rule.
X(s) = D_x / D = 0/-3s^2 +1 = 0
Y(s) = D_y / D = -3s^2 + 1 / s(-3s^2 + 1) = 1/s
then take the laplace transforms and I get x(t)=0 and y(t)=1

Can someone please tell me if this is right? Thanks.

 

[HallsofIvy]

That's not at all the way I would do the problem (I detest "Laplace transform") but that's exactly what I got as the answer: x(t)= 0 and y(1)= 1. Of course, you could have checked that yourself. Since x and y are constants, there derivatives are 0 and the equations reduce to 0+ 2(0)+ 0= 0 and 0- 0+ 1= 1.

 

[tacman]

Your solution looks correct to me. I also verified it by solving the system of equations using the method of undetermined coefficients and got the same answer. However, I would like to point out a small mistake in your calculations. When finding D_y, the second term in the second row should be -s+1 instead of -s. Other than that, your solution is correct. Good job!