LaPlace Transform problem?

Question

Find the solution of the initial value problem by the use of Laplace Transforms. Show all steps.
x1'= - x1 - 3 x2 + 3 x3 x1(0) = -1
x2' = -4 x2 + 3 x3 x2(0) = 1
x3' = -6 x2 + 5 x3 x3(0) = 0

Does anyone know how to do this or at least start it?

x1 = x sub 1
x2 = x sub 2
x3 = x sub 3

Answer 1

Use the property
L{f'(t)} = s*F(s) - f(0)

Now I prefer to use x = x1, y = x2, z = x3
L{x(t)} = X(s) ie lower case to upper case

Take Laplace transform of each of the equations applying the equation above:

sX + 1 = -X - 3Y + 3Z
sY - 1 = -4Y + 3Z
sZ = -6Y + 5Z

Now you have 3 linear equations to solve simultaneously in terms of s. Then you have to take the inverse transforms. It's a bit of really messy algebra involved.

I'll do one of them for you.
Z(s) = -6 / (s^2 - s + 2)
= -6 / [(s-1/2)^2 + 7/4 ]
Taking the inverse transform
z(t) = -12/sqrt(7) * exp(t/2) * sin[t*sqrt(7)/2]

Answer 2

are there no exponents???