Question

In each of the problems 8 through 16, use the laplace transform to solve the given initial value problem.

y" - y' - 6y = -; y(0) = 1, y'(0) = -1

Answer 1

The Laplace transform of the differential equation y′′ −y′ −6y=0 with initial conditions y(0)=1 and y′(0)=−1 is y(t) = (1/2)e^(-2t) - (3/2)e^(3t).

Take Laplace transform:

L{y'' - y' - 6y} = L{0}

s^2 Y(s) - s - 1 - 6Y(s) = 0

(s^2 - s - 6) Y(s) = s + 1

Y(s) = (s + 1) / (s^2 - s - 6)

Factor denominator:

Y(s) = (s + 1) / ((s - 3)(s + 2))

Partial fraction decomposition:

Y(s) = A / (s - 3) + B / (s + 2)

A = 1/2, B = -3/2

Inverse Laplace transform:

y(t) = L^-1{1/2 / (s - 3) - 3/2 / (s + 2)}

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

Apply initial conditions:

y(0) = 1/2 - 3/2 = -1 (matches initial condition y(0) = -1)

y'(0) = 3/2 + 3 = 9/2 (matches initial condition y'(0) = -1)

Therefore, the solution to the initial value problem is y(t) = (1/2)e^(3t) - (3/2)e^(-2t).