Solve the following initial value problem using laplace tran

Solve the following initial value problem using laplace transform:

y\'\'+4y=4, y(0)=0, y\'(0)=1

Solution

y\'\' + 4y = 4
Let\'s \"laplace-transform\" both sides. We get:
L(y\'\' + 4y) = L(4)
L(y\'\') + 4 L(y) = L(4)

Remember:
* L(cst) = cst / s so L(4) = 4/s
* L(y\'\') = s^2 L(y) - s y(0) - y\'(0) . So the problem has to specify y(0) and y\'(0) in order to complete the exercise. For now, we have y(0) = 0 and y\'(0) = 1
We get L(y\'\') = s^2 L(y) - 1

Go back and replace in the equation:
s^2 L(y) - 1 + 4L(y) = 4/s
(s^2 + 4) L(y) = 4/s + 1

L(y) = 4/(s(s^2 + 4)) + 1/(s^2 + 4)

Let\'s decompose 4/(s(s^2 + 4) = A/s + (Bs + C) / (s^2 + 4)
= [ A(s^2 + 4) + s(Bs+C) ] / (s(s^2 + 4)) = [(A+B)s^2 + Cs + 4A] / (s(s^2 + 4))
Identifying with 4 / (s(s^2 + 4)) we get A+B = 0 AND C = 0 And 4A = 4
So A = 1, B = -1, and C = 0
So 4/(s(s^2 + 4)) = 1/s - s/(s^2 + 4)

Going back to the expression of L(y) we get

L(y) = 1/s - s/(s^2 + 4) + 1/(s^2 + 4)
= 1/s - s/(s^2 + 4) + (1/2)*(2/(s^2 + 4))

Now we apply L^-1 to get to get y:
y =