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| The [[Laplace transform]] is a powerful [[integral transform]] used to switch a function from the [[time domain]] to the [[Laplace transform#s-Domain equivalent circuits and impedances|s-domain]]. The Laplace transform can be used in some cases to solve [[linear differential equation]]s with given [[Initial value problem|initial conditions]].
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| First consider the following property of the Laplace transform:
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| :<math>\mathcal{L}\{f'\}=s\mathcal{L}\{f\}-f(0)</math>
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| :<math>\mathcal{L}\{f''\}=s^2\mathcal{L}\{f\}-sf(0)-f'(0)</math>
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| One by [[Mathematical induction|induction]] can prove that
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| :<math>\mathcal{L}\{f^{(n)}\}=s^n\mathcal{L}\{f\}-\sum_{i=1}^{n}s^{n-i}f^{(i-1)}(0)</math>
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| Now we consider the following differential equation:
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| :<math>\sum_{i=0}^{n}a_if^{(i)}(t)=\phi(t)</math>
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| with given initial conditions
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| :<math>f^{(i)}(0)=c_i</math>
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| Using the [[linearity]] of the Laplace transform it is equivalent to rewrite the equation as
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| :<math>\sum_{i=0}^{n}a_i\mathcal{L}\{f^{(i)}(t)\}=\mathcal{L}\{\phi(t)\}</math>
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| obtaining
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| :<math>\mathcal{L}\{f(t)\}\sum_{i=0}^{n}a_is^i-\sum_{i=1}^{n}\sum_{j=1}^{i}a_is^{i-j}f^{(j-1)}(0)=\mathcal{L}\{\phi(t)\}</math>
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| Solving the equation for <math> \mathcal{L}\{f(t)\}</math> and substituting <math>f^{(i)}(0)</math> with <math>c_i</math> one obtains
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| :<math>\mathcal{L}\{f(t)\}=\frac{\mathcal{L}\{\phi(t)\}+\sum_{i=1}^{n}\sum_{j=1}^{i}a_is^{i-j}c_{j-1}}{\sum_{i=0}^{n}a_is^i}</math>
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| The solution for ''f''(''t'') is obtained by applying the [[inverse Laplace transform]] to <math>\mathcal{L}\{f(t)\}.</math>
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| Note that if the initial conditions are all zero, i.e.
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| :<math>f^{(i)}(0)=c_i=0\quad\forall i\in\{0,1,2,...\ n\}</math>
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| then the formula simplifies to
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| :<math>f(t)=\mathcal{L}^{-1}\left\{{\mathcal{L}\{\phi(t)\}\over\sum_{i=0}^{n}a_is^i}\right\}</math>
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| ==An example==
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| We want to solve
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| :<math>f''(t)+4f(t)=\sin(2t)</math>
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| with initial conditions ''f''(0) = 0 and ''f′''(0)=0.
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| We note that
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| :<math>\phi(t)=\sin(2t)</math>
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| and we get
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| :<math>\mathcal{L}\{\phi(t)\}=\frac{2}{s^2+4}</math>
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| The equation is then equivalent to
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| :<math>s^2\mathcal{L}\{f(t)\}-sf(0)-f'(0)+4\mathcal{L}\{f(t)\}=\mathcal{L}\{\phi(t)\}</math>
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| We deduce
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| :<math>\mathcal{L}\{f(t)\}=\frac{2}{(s^2+4)^2}</math> | |
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| Now we apply the Laplace inverse transform to get
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| :<math>f(t)=\frac{1}{8}\sin(2t)-\frac{t}{4}\cos(2t)</math>
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| ==Bibliography==
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| * A. D. Polyanin, ''Handbook of Linear Partial Differential Equations for Engineers and Scientists'', Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 1-58488-299-9
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| [[Category:Integral transforms]]
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| [[Category:Differential equations]]
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| [[Category:Differential calculus]]
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| [[Category:Ordinary differential equations|Ordinary differential equations]]
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