Can someone help me with this Laplace Transform please?

Question

Use the Laplace Transform to solve the given system of differential equations.

dx/dt = -x + y
dy/dt = 2x
x(0)=0, y(0)=1


THANKS!

Answer 1

The solution of the system, after applying Laplace Transform, solving for the system and then taking the Inverse Laplace Transform is

x = 1/3 e^(-2t) ( -1 + e^(3t) )

y = 1/3 e^(-2t) + 2/3e^t

Answer 2

dx/dt = y-x, dy/dt = 2x, x(0) = 0, y(0) = 1.

In the Laplace domain:
s L{x} - x(0) = L{y} - L{x}
s L{y} - y(0) = 2 L{x}

(s+1) L{x} = L{y}
s L{y} = 1 + 2 L{x}
Substitute the L{y} from the top eqn in
place of the L{y} in the bottom equation.
s(s+1) L{x} = 1 + 2 L{x}
(s^2 + s - 2} L(x) = 1
L{x} = 1/[(s+2)(s-1)]

Now use "partial fractions":
1/[(s+2)(s-1)] = A/(s+2) + B/(s-1)
1 = As - A + Bs + 2B
A + B = 0, 2B - A = 1 =>
B = 1/3, A = -1/3.

L{x} = (1/3)/(s-1) - (1/3)/(s+2)
x(t) = (1/3)e^t - (1/3)e^(-2t)
dy/dt = (2/3)e^t - (2/3)e^(-2t)
y = (2/3)e^t - (1/3)e^(-2t) + C
and the initial condition shows that C = 2/3.