Solving Differential Equation w/ Laplace Transform

[Neoon]

Hi all,

Just a small question:
I came across a problem to solve a differential equation using Laplace transform.

I solved the major part but only still a part where I had to inverse transform the following experssion:

(exp(-s))/(s^2)

I looked in Laplace Transform Tables but did not find it.

Could you please help?

 

[tpm]

think i got it..

you're looking for the integral..

$$(2\pi i)^{-1} \oint ds exp(-st-s)/s^{2}$$

now using 'Residue theorem' you get...

$$f(t)=0$$ iff t<1 and f(t)=t/2 [/tex] iff t>1.


depending on if you take the semi-circle (-T,0) or (0,T) with T--->oo , there is a double pole at s=0

 

[tacman]

Hi there,

The expression (exp(-s))/(s^2) can be solved using the Laplace transform by using the property of differentiation in the Laplace domain. This property states that if the Laplace transform of a function f(t) is F(s), then the Laplace transform of f'(t) is sF(s) - f(0). In this case, the function f(t) is exp(-s) and its derivative is -exp(-s).

So, we can rewrite the expression as follows:

(exp(-s))/(s^2) = (-exp(-s))/s * (1/s)

Using the property of differentiation, we get:

(-exp(-s))/s = L{f'(t)} = sF(s) - f(0)

Therefore, we can rewrite the expression as:

(exp(-s))/(s^2) = [sF(s) - f(0)] * (1/s) = F(s) - (f(0)/s)

Now, we can use the inverse Laplace transform to find the solution for F(s). The inverse Laplace transform of sF(s) is f'(t), so we get:

F(s) = f'(t) + (f(0)/s)

Substituting this in the previous expression, we get:

(exp(-s))/(s^2) = f'(t) + (f(0)/s) - (f(0)/(s^2))

Finally, we can take the inverse Laplace transform of both sides to get the solution for f(t):

f(t) = L{f'(t)} + L{(f(0)/s)} - L{(f(0)/(s^2))}

The first term is simply f(t), the second term is f(0), and the third term is (f(0)t). So, the final solution is:

f(t) = exp(-t) + f(0) - (f(0)t)

I hope this helps! Let me know if you have any further questions.