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Laplace transform applied to differential equations - Wikipedia, the free encyclopedia

Laplace transform applied to differential equations

From Wikipedia, the free encyclopedia

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=1}s^{i-j}f^{(j-1)}(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)\}

[edit] 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 that

\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)

[edit] Bibliography

  • 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

[edit] External links

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