Answer:
y(t) = 2e^(-3t) + 5te^(-3t).
Explanation:
1. Take the Laplace transform of the differential equation:
The Laplace transform of y' is sY(s) - y(0), where Y(s) represents the Laplace transform of y.
Applying the Laplace transform to the given differential equation, we get:
sY(s) - y(0) + 3Y(s) = 1/(s + 3)
2. Substitute the initial condition into the equation:
Since y(0) = 2, we substitute y(0) = 2 into the equation obtained in step 1:
sY(s) - 2 + 3Y(s) = 1/(s + 3)
3. Solve the equation for Y(s):
Combine like terms and isolate Y(s) on one side:
(s + 3)Y(s) = 1/(s + 3) + 2
(s + 3)Y(s) = (1 + 2(s + 3))/(s + 3)
(s + 3)Y(s) = (2s + 7)/(s + 3)
Divide both sides by (s + 3):
Y(s) = (2s + 7)/(s + 3)^2
4. Simplify Y(s) if possible:
Y(s) can be further simplified by expanding the numerator:
Y(s) = (2s + 7)/(s^2 + 6s + 9)
5. Inverse Laplace transform to find y(t):
To find y(t), we need to take the inverse Laplace transform of Y(s). The inverse Laplace transform of Y(s) is denoted as y(t).
Using partial fraction decomposition or the table of Laplace transforms, we can find that the inverse Laplace transform of Y(s) is:
y(t) = 2e^(-3t) + 5te^(-3t)
Therefore, the solution to the initial-value problem is y(t) = 2e^(-3t) + 5te^(-3t).