inverse laplace transform of {5/(s-4)^3}
L{e^at f(t)} = F(s-a)
so,
L{e^4t f(t)} = F(s-4)
now, what is L^-1(1/s^3) ?
To find the inverse Laplace transform of 5/(s-4)^3, we will use the formula for the inverse Laplace transform of a function in the form of F(s)/(s-a)^n.
The formula states that if F(s) = L^-1[F(s)] is the inverse Laplace transform of F(s), then the inverse Laplace transform of F(s)/(s-a)^n is given by:
L^-1[F(s)/(s-a)^n] = (d^n/ds^n)[e^(as)F(s)]
In our case, F(s) = 5, a = 4, and n = 3. So, we can write the inverse Laplace transform of 5/(s-4)^3 as:
L^-1[5/(s-4)^3] = (d^3/ds^3)[e^(4s) * 5]
Differentiating the term e^(4s) * 5 three times with respect to s will give us the inverse Laplace transform.
First, let's find the first derivative:
(d/ds)[e^(4s) * 5] = 4e^(4s) * 5
Second derivative:
(d^2/ds^2)[e^(4s) * 5] = d/ds[4e^(4s) * 5] = 4*4e^(4s) * 5
Third derivative:
(d^3/ds^3)[e^(4s) * 5] = d/ds[4*4e^(4s) * 5] = 4*4*4e^(4s) * 5
Simplifying this expression, we get:
L^-1[5/(s-4)^3] = 4*4*4*5e^(4s)
Finally, simplifying further:
L^-1[5/(s-4)^3] = 320e^(4s)
Therefore, the inverse Laplace transform of 5/(s-4)^3 is 320e^(4s).
To find the inverse Laplace transform of the given function, \(\frac{5}{(s-4)^3}\), we can use the property of Laplace transforms that states:
\[\mathcal{L}^{-1}\left\{\frac{1}{(s-a)^n}\right\} = \frac{t^{n-1}e^{at}}{(n-1)!}\]
Applying this property to our function, where \(a = 4\) and \(n = 3\), we have:
\[\mathcal{L}^{-1}\left\{\frac{5}{(s-4)^3}\right\} = \frac{t^{3-1}e^{4t}}{(3-1)!} = \frac{t^2e^{4t}}{2}\]
Therefore, the inverse Laplace transform of \(\frac{5}{(s-4)^3}\) is \(\frac{t^2e^{4t}}{2}\).