laplace transform applied to differential equations

The use of Laplace transform makes it much easier to solve linear differential equations with given initial conditions. First consider the following relations:

\mathcal{L}\{f'\}

   = s \mathcal{L}\{f\} - f(0)

\mathcal{L}\{f''\}

   = s^2 \mathcal{L}\{f\} - s f(0) - f'(0)

\mathcal{L}\{f^{(n)}\}

   = s^n \mathcal{L}\{f\} - \Sigma_{i = 1}^{n}s^{n - i}f^{(i - 1)}(0)

Suppose we want to solve the given differential equation:

\sum^n_{i=0}a_if^{(i)}(t)=\phi(t)

This equation is equivalent to

\sum^n_{i=0}a_i\mathcal{L}\{f^{(i)}(t)\}=\mathcal{L}\{\phi(t)\}

which is equivalent to

\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}

note that the

f^{(k)}(0)

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

\mathcal{L}\{f(t)\}

An example

We want to solve :

f^{(2)}(t)+4f(t)=\sin(2t) \,\!

with initial conditions f(0) = 0 and f ′(0)=0 we note :

\phi(t)=\sin(2t) \,\!

and we get :

\mathcal{L}\{\phi(t)\}=\frac{2}{s^2+4}

so this is equivalent to :

s^2\mathcal{L}\{f(t)\}-sf(0)-f^{(1)}(0)+4\mathcal{L}\{f(t)\}=\mathcal{L}\{\phi(t)\}

we deduce :

\mathcal{L}\{f(t)\}=\frac{2}{(s^2+4)^2}

So we apply the Laplace inverse transform and get

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

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