Laplace transform: x′′ + 4x = cos 2t, x(0) = 1, x′ (0) = −2

[joshderise]

Using Laplace transforms solve the following initial-value problem
x′′ + 4x = cos 2t, x(0) = 1, x′ (0) = −2.

im not sure if im on the right track .

s^2.X(s) - s +2 + 4.X(s)=s/(s^2+4)

X(s) = (s+s^3+4s-2s^2-8)/(s^2+4)

with thanks , josh

 

[HallsofIvy]

Using Laplace transforms solve the following initial-value problem
x′′ + 4x = cos 2t, x(0) = 1, x′ (0) = −2.

im not sure if im on the right track .

s^2.X(s) - s +2 + 4.X(s)=s/(s^2+4)

X(s) = (s+s^3+4s-2s^2-8)/(s^2+4)

with thanks , josh

Looks good so far. Now, since that is an "improper fraction", the numerator, s^3- 2s^2+ 5s- 8, has higher degree that the numerator, s^2+ 4, do the division to get a simple polynomial plus a proper fraction.