Find the Particular Solution of the following system using the Laplace Transform method:

Question

$x' = x - 2y$

$y' = 3x - 4y$

$x(0) = 1$

$y(0) = 1$

How would the Laplace Transform method differ from the usual methods? My first thought is that I should equate the two equations using $x$ or $y$.

Answer 1

Laplace transforming we have

$$ sX = X-2Y+x_0\\ sY = 3X-4Y + y_0 $$

Now solving for $X,Y$

$$ X = \frac{(s+4) x_0-2 y_0}{s^2+3 s+2},Y = \frac{(s-1) y_0+3 x_0}{s^2+3 s+2} $$

so

$$ x = e^{-2 t} \left(e^t (3 x_0-2 y_0)-2 x_0+2 y_0\right)u(t)\\ y = e^{-2 t} \left(3 \left(e^t-1\right) x_0+\left(3-2 e^t\right) y_0\right)u(t) $$

with $u(t)$ the unit step function

etc.