Laplace transform applied to differential equations
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The use of Laplace transform makes it much easier to solve linear differential equations with given initial conditions.
First consider the following relations:
- <math>\mathcal{L}\{f'\}
= s \mathcal{L}\{f\} - f(0)<math>
- <math>\mathcal{L}\{f''\}
= s^2 \mathcal{L}\{f\} - s f(0) - f'(0)<math>
- <math>\mathcal{L}\{f^{(n)}\}
= s^n \mathcal{L}\{f\} - \Sigma_{i = 1}^{n}s^{n - i}f^{(i - 1)}(0)<math>
Suppose we want to solve the given differential equation:
- <math>\sum^n_{i=0}a_if^{(i)}(t)=\phi(t)<math>
This equation is equivalent to
- <math>\sum^n_{i=0}a_i\mathcal{L}\{f^{(i)}(t)\}=\mathcal{L}\{\phi(t)\}<math>
which is equivalent to
<math>\mathcal{L}\{f(t)\}={\mathcal{L}\{\phi(t)\}+\sum^n_{i=0}a_i\sum^i_{j=0}s^{i-j}f^{(j-i)}(0) \over \sum^n_{i=0}a_is^i}<math>
note that the <math>f^{(k)}(0)<math> are initial conditions.
Then all we need to get f(t) is to apply the Laplace inverse transform to <math>\mathcal{L}\{f(t)\}<math>
An example
We want to solve :
- <math>f^{(2)}(t)+4f(t)=\sin(2t) \,\!<math>
with initial conditions f(0) = 0 and f ′(0)=0
we note :
- <math>\phi(t)=\sin(2t) \,\!<math>
and we get :
- <math>\mathcal{L}\{\phi(t)\}=\frac{2}{s^2+4}<math>
so this is equivalent to :
- <math>s^2\mathcal{L}\{f(t)\}-sf(0)-f^{(1)}(0)+4\mathcal{L}\{f(t)\}=\mathcal{L}\{\phi(t)\}<math>
we deduce :
- <math>\mathcal{L}\{f(t)\}=\frac{2}{(s^2+4)^2}<math>
So we apply the Laplace inverse transform and get
- <math>f(t)=\frac{1}{8}\sin(2t)-\frac{t}{4}\cos(2t) <math>
Bibliography
- A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, 2002.
fr:Application de la transformée de Laplace aux équations différentielles