what is the laplace transform of
5/2.e^t-3/2.e^-t
L{e^t} = 1/(s-1)
L{e^-t} = 1/(s+a)
L is a linear transform, so
L{5/2.e^t-3/2.e^-t} = 5/2 L{e^t} - 3/2 L{e^-t}
= 5/2 * 1/(s-1) - 3/2 * 1/(s+1)
= (s+4)/(s^2-1)
compute the inverse laplace transform 2S^2+13S+5/(s+3)(s-1)^2
i got 3e^t+5e^t+3e^t, not sure it correct
To find the Laplace transform of the given function, you can use the definition of the Laplace transform and some basic properties.
The Laplace transform of a function f(t) is given by the integral:
F(s) = L{f(t)} = ∫[0, ∞] (f(t) * e^(-st)) dt,
where s is a complex number (frequency domain variable) and L denotes the Laplace transform operator.
In this case, the function is f(t) = (5/2)e^t - (3/2)e^(-t).
To find its Laplace transform F(s), we will need to split it into two separate terms:
F(s) = L{ (5/2)e^t - (3/2)e^(-t) }.
Applying linearity property of the Laplace transform, we can compute the Laplace transform of each term separately.
First, let's find the Laplace transform of (5/2)e^t.
Using the property L{e^(at)} = 1/(s-a), we have:
L{ (5/2)e^t } = (5/2) * L{ e^t } = (5/2) * 1 / (s - 1).
Similarly, let's find the Laplace transform of (3/2)e^(-t).
Using the property L{e^(at)} = 1/(s-a), we have:
L{ (3/2)e^(-t) } = (3/2) * L{ e^(-t) } = (3/2) * 1 / (s + 1).
Now we can combine the two terms to get the Laplace transform of the entire function:
F(s) = (5/2) * 1 / (s - 1) - (3/2) * 1 / (s + 1).
Thus, the Laplace transform of (5/2)e^t - (3/2)e^(-t) is given by:
F(s) = (5/2) * 1 / (s - 1) - (3/2) * 1 / (s + 1).