ice007 10 Posted October 20, 2008 Share Posted October 20, 2008 Hi All, I really need to understand laplace transform as i have an assignment on it. Its really confusing for me can anyone assist me on the following questions. Please help me to understand s^4 – 6s^3 + 18s^2 / 3s^6 ac + bs / s^2 + a^2 s + 5 / s2 - 6s + 25 and a differential equation using laplace tranform f'(t) + f(t) = e^-t, f(0) = 0 Thank you very much Link to post Share on other sites

Dave 251 Posted October 20, 2008 Share Posted October 20, 2008 Essentially the Laplace transform is just a function itself; it takes one function, then integrates it to get another: [math]\mathcal{L}[f](s) = \int_0^{\infty} e^{-st} f(t) \, dt[/math] So, to calculate the Laplace transform of a function f, one simply calculates the integral on the right hand side to get another function in terms of s. For example, for the function [math]f(t) = e^{-t}[/math], we have that the Laplace transform is: [math]\mathcal{L}[f](s) = \int_0^{\infty} e^{-st} e^{-t} \, dt = \int_0^\infty e^{-(s+1)t} \, dt[/math] so [math]\mathcal{L}[f](s) = \left[ \frac{e^{-(s+1)t}}{s+1} \right]_\infty^0 = \frac{1}{s+1}[/math]. The Laplace transform is generally useful for finding the solutions to various ODEs, the reason being that: [math]\mathcal{L}[f'](s) = s\mathcal{L}[f](s) - f(0)[/math] You can prove this by writing out the left hand side, and then using integration by parts. So, for the equation that you have to solve, we can apply the Laplace transform to both sides: [math]\mathcal{L}[f'(t) + f(t)](s) = \mathcal{L}[e^{-t}](s)[/math] Now the Laplace transform is linear (easy to prove) so you can split up the left hand side. Also we use the transform I proved earlier for [imath]e^{-t}[/imath]: [math]\mathcal{L}[f'](s) + \mathcal[f](t)(s) = \frac{1}{s+1}[/math] Use the property of the Laplace transform of a derivative to get: [math]s\mathcal{L}[f](s) - f(0) + \mathcal{L}[f](s) = \frac{1}{s+1}[/math] and finally we simplify everything down, using the fact that [imath]f(0) = 0[/imath]: [math](s+1)\mathcal{L}[f](s) = \frac{1}{s+1} \Rightarrow \mathcal{L}[f](s) = \frac{1}{(s+1)^2}[/math] So the problem boils down to: can we find an f whose Laplace transform is [imath](s+1)^{-2}[/imath]? Or, in other words, what is the inverse Laplace transform of [imath](s+1)^{-2}[/imath]? Unfortunately the inverse Laplace transform is difficult to calculate, so basically you have to look it up in a table. In our case, we're lucky and this is a common inverse: [math]f(t) = te^{-t}[/math] You can find the list on Wikipedia - just search for Laplace Transform. If you put this into the original equation, you can see that this is indeed a solution to the ODE problem. Hope this helps. Link to post Share on other sites

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