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

Answer: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 equationL{y''(t)} = s^2Y(s) - sy(0) - y'(0) = (s^2)Y(s) - sL{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) = 0Y(s)(s^2 + 3s + 2) = sY(s)(s+2)(s+1) = sY(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 = 2B = s/(s+2) when s = -1, so B = -1SoY(s) = 2/(s+2) - 1/(s+1)Doing the inverse Laplace transformy(t) = 2e^(-2t) - e^(-t)