what is the Laplace Transform for e^(-2t) u(t-1) Solution Laplace transform of(t

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Laplace transform of(t^1/2 erf(squrt t).I wnt knw the working and how to arrive at the final answer. ? Note that f(t) = t^(1/2) erf(vt) = t^(1/2) * (2/vp) ?(0 to vt) e^(-u^2) du f '(t) = (2/vp) {(1/2)t^(-1/2) * ?(0 to vt) e^(-u^2) du + t^(1/2) * [e^(-t) * (1/2)t^(-1/2)]} So, 2t f '(t) - f(t) = (2/vp) te^(-t) Moreover, we also have the initial condition f(0) = 0. Apply L to both sides: -2 (d/ds) (s F(s) - 0) - F(s) = (2/vp) /(s+1)^2 ==> -2(F(s) + s F'(s)) - F(s) = (2/vp) /(s+1)^2 ==> F'(s) + (3/(2s)) F(s) = (-1/vp) /[s(s+1)^2]. This is a first order linear DE in F(s) with integrating factor s^(3/2): s^(3/2) F'(s) + (3/2) s^(1/2) F(s) = (-1/vp) s^(1/2)/(s+1)^2 ==> (d/ds) [s^(3/2) F(s)] = (-1/vp) s^(1/2)/(s+1)^2 Integrate both sides: s^(3/2) F(s) = (-1/vp) ? s^(1/2) ds/(s+1)^2 s^(3/2) F(s) = (-1/vp) [arctan(vs) - vs /(s+1)] + C. Link: http://www.wolframalpha.com/input/?i=int… Solving for F(s): F(s) = (-1/vp) [s^(-3/2) arctan(vs) - 1/(s(s+1))] + C .......= (1/vp) [-s^(-3/2) arctan(vs) + 1/(s(s+1))] + C. To find C, use the Initial Value Theorem: lim(t?0) f(t) = lim(s?8) s F(s) Since lim(t?0) f(t) = f(0) = 0, we have lim(s?8) s F(s) = 0. Thus, lim(s?8) s F(s) = 0 ==> lim(s?8) (1/vp) [-s^(-1/2) arctan(vs) + 1/(s+1)] + Cs = 0. Since lim(s?8) (1/vp) [-s^(-1/2) arctan(vs) + 1/(s+1)] = 0, we require that C = 0. ------------------- In summary, L{t^(1/2) erf(vt)} = (1/vp) [-s^(-3/2) arctan(vs) + 1/(s(s+1))].

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