Use the Laplace transform to solve the initial value problem"3y + 2y=0, y (0) = 1, y (0) = 0.

Y (s) = 2 / (s+2) - 1 / (s+1)

y (t) = 2e^ (-2t) - e^ (-t)

Step-by-step explanation:

Let's solve y'' + 3y' + 2y = 0, by the Laplace transform.

So, the first step is to find the Laplace transform of each component of the differential equation

L{y'' (t) } = s^2Y (s) - sy (0) - y' (0) = (s^2) Y (s) - s

L{3y' (t) } = 3L{y' (t) } = 3 (sY (s) - y (0)) = 3sY (s)

L{2y (t) } = 2L{y (t) } = 2Y (s)

So now we have to solve the following equation:

(s^2) Y (s) - s + 3sY (s) + 2Y (s) = 0

Y (s) (s^2 + 3s + 2) = s

Y (s) (s+2) (s+1) = s

Y (s) = (s / (s+2) (s+1))

Now to find y (t) and solve the initial value problem, we need to do the inverse Laplace transform of Y (s), that we can do the following way:

Y (s) = (s / (s+2) (s+1)) = (A / (s+2)) + (B / (s+1))

where A = s / (s+1) when s = - 2, so A = 2

B = s / (s+2) when s = - 1, so B = - 1

So

Y (s) = 2 / (s+2) - 1 / (s+1)

Doing the inverse Laplace transform

y (t) = 2e^ (-2t) - e^ (-t)