Laplace TransformsDate: 06/21/97 at 01:35:22 From: Kenneth Subject: Laplace Transform Hi! I have a little trouble understanding the linearity of the Laplace transform. Defined as "the Laplace transform is a linear operation; that is, for any functions f(t) and g(t) whose Laplace transforms exist and any constants a and b." I do not understand the meaning. Could you rephrase this so it's easier to understand? What's the point of this theorem? If f(t) = tsinwt, L(f) = ? Solution: f(0) = 0 f'(t) = sin wt + wt cos wt, f'(0)=0 What's next? Thank you very much for your time and help, Kenneth Date: 06/21/97 at 11:21:33 From: Doctor Anthony Subject: Re: Laplace Transform A linear operator is similar to, say, differentiation or integration; that is: d/dx[f(x) + g(x)] = f'(x) + g'(x) and d/dx(a*f(x) = a*f'(x) A non-linear operator would be squaring, i.e., (a+b)^2 is NOT equal to a^2 + b^2. To find the Laplace transform of xsin(wx) we evaluate: L[xsin(wx)] = INT(0 to inf.)[e^(-px)xsin(wx)dx] 2wp = ------------ (p^2 + w^2)^2 To solve f'(x) = sin(wx) + wxcos(wx), we write down the transforms term by term, to get: -y(0) + p*y(p) = w/(p^2+w^2) + w[(p^2-w^2)/(p^2+w^2)^2 and y(0) = 0 w w(p^2-w^2) y(p) = --------- + -------------- p(p^2+w^2) p(p^2+w^2)^2 At this stage we use partial fractions to separate the terms in the denominators. I will work though the first one: 1 A Bp + C -------- = ------ + -------- p(p^2+w^2) p p^2 + w^2 1 = A(p^2 + w^2) + Bp^2 + Cp p = 0 gives 1 = Aw^2, so A = 1/w^2. Also A+B = 0, so B = -1/w^2 and C = 0 because there is no term in p on the lefthand side. 1 1 1 ----------- = --------- - ------------- p(p^2+w^2) w^2*p w^2(p^2+w^2) and w 1 w --------- = ------- - ------------- p(p^2+w^2) w*p w^2(p^2+w^2) We now use the inverse transforms to write down the result: = 1/w - (1/w^2)sin(wx) The second fraction is: wp^2 - w^3 wp w^3 ------------- = ---------- - ----------- p(p^2+w^2)^2 (p^2+w^2)^2 p(p^2+w^2)^2 The first of these is (1/2)x*sin(wx) and the second one requires a further application of partial fractions. 1 A Bp + C Dp + E -------- = ------ + -------- + ---------- p(p^2+w^2)^2 p (p^2+w^2) (p^2+w^2)^2 Some rather tedious algebra and a table of inverse transforms will allow you to complete the solution of the differential equation. The use of these transforms depends very much on not having 'p' functions with unknown inverses. It is a good idea to collect as many inverses as you can find from a variety of textbooks. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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