Hi

The following question i am having trouble solving

Use Laplace Transforms to solve each system of differential equations

$\displaystyle \frac{dx}{dt} = -2y+1$

$\displaystyle \frac{dy}{dt} = 2x-t$

given x(0) = -1, y(0) = 1

This is what i have done:

$\displaystyle sX+2Y = 0$

$\displaystyle sY-2X=1-\frac{1}{s^2}$

I used cramers rule to find X and Y

$\displaystyle Y = \frac{s^2-1}{s(s^2+4)}$

$\displaystyle Y = \frac{s^2}{s(s^2+4)}-\frac{1}{s(s^2+4)}$

$\displaystyle Y = cos(2t)-\frac{1}{4}+\frac{cos(2t)}{4}$

$\displaystyle X = \frac{-2s^2+2}{s^2(s^2+4)}$

$\displaystyle X = \frac{-2s^2}{s^2(s^2+4)}+\frac{2}{s^2(s^2+4)}$

$\displaystyle X = -sin(2t) + \frac{2t}{4}-\frac{2sin(2t)}{8}$

Answer:

$\displaystyle X = \frac{1}{2}(t - \frac{3sin(2t)}{2})-cos(2t)$

$\displaystyle Y = \frac{1}{4}(1+3cos(2t))-sin(2t)$

P.S