Laplace Transform Applied To Differential Equations

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)\}

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)\}=rac{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)\}=rac{2}{(s^2+4)^2}

So we apply the Laplace inverse transform and get

:

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

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

EXTERNAL LINKS

	CATEGORIES ABOUT LAPLACE TRANSFORM APPLIED TO DIFFERENTIAL EQUATIONS

	integral transforms

	differential equations

	differential calculus

	ordinary differential equations

Information About