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>

fr:Application de la transformée de Laplace aux équations différentielles