Laplace transform applied to differential equations
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The use of Laplace transform makes it much easier to solve linear differential equations.

First consider the folowing relations :

Suppose we want to solve the given differential equation:

this equation is equivalent to :

which is equivalent to :

note that the are initial conditions.

Then all we need to get f(t) is to apply the Laplace inverse transform to

An example

We want to solve :

with initial conditions f(0) = 0 and f ′(0)=0

we note :

and we get :

so this is equivalent to :

we deduce :

So we apply the Laplace inverse transform and get

$f(t)=\backslash \backslash frac\{1\}\{8\}\backslash \backslash sin(2t)-\backslash \backslash frac\{t\}\{4\}\backslash \backslash cos(2t)$

The original article can be found at: http://en.wikipedia.org/wiki/Laplace_transform_applied_to_differential_equations.
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