what is the laplace transform of cost+4sint
You will find the answer here.
http://en.wikipedia.org/wiki/Laplace_transform
Set w (omega) = 1.
Add the Laplace transforms of the two terms.
I get (1 + 4s)/(s^2 + 1)
Hmm. I get (s+4)/(s^2+1)
To find the Laplace transform of a given function, we can use the linearity property of the Laplace transform.
The linearity property states that for any constants a and b, and any functions f(t) and g(t), the Laplace transform of the linear combination af(t) + bg(t) is given by a times the Laplace transform of f(t) plus b times the Laplace transform of g(t).
Let's apply this property to find the Laplace transform of the function cos(t) + 4sin(t):
1. Start by finding the Laplace transform of the function cos(t).
The Laplace transform of cos(t) can be obtained using a table of Laplace transforms:
L{cos(t)} = s / (s^2 + 1)
2. Next, find the Laplace transform of the function sin(t).
The Laplace transform of sin(t) can also be found in the table:
L{sin(t)} = 1 / (s^2 + 1)
3. Use the linearity property to find the Laplace transform of cos(t) + 4sin(t):
L{cos(t) + 4sin(t)} = L{cos(t)} + 4 * L{sin(t)}
= (s / (s^2 + 1)) + 4 * (1 / (s^2 + 1))
Combining these terms, we get the Laplace transform of cos(t) + 4sin(t) as:
L{cos(t) + 4sin(t)} = (s + 4) / (s^2 + 1)
Therefore, the Laplace transform of cos(t) + 4sin(t) is (s + 4) / (s^2 + 1).