Method for Solving Linear Differential Equations Using the Laplace Transform

In the vast and intricate landscape of mathematics , the Laplace transform stands as a formidable integral transform , serving as a bridge that transports functions from the time domain to the s-domain . This transformation is not merely a mathematical curiosity; it is a powerful tool that can be wielded to solve linear differential equations with given initial conditions . The elegance of the Laplace transform lies in its ability to convert complex differential equations into algebraic equations, which are often easier to manipulate and solve.

Approach

To understand the methodology, one must first acquaint themselves with a fundamental property of the Laplace transform. Consider the following properties:

For the first derivative of a function ( f(t) ):

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

For the second derivative:

[ {\mathcal {L}}{f’’} = s^{2}{\mathcal {L}}{f} - sf(0) - f’(0) ]

These properties can be generalized using mathematical induction to the nth derivative:

[ {\mathcal {L}}{f^{(n)}} = s^{n}{\mathcal {L}}{f} - \sum _{i=1}^{n}s^{n-i}f^{(i-1)}(0) ]

Now, let’s consider the following linear differential equation:

[ \sum {i=0}^{n}a{i}f^{(i)}(t) = \phi (t) ]

with given initial conditions:

[ f^{(i)}(0) = c_{i} ]

Using the linearity of the Laplace transform, we can rewrite the equation as:

[ \sum {i=0}^{n}a{i}{\mathcal {L}}{f^{(i)}(t)} = {\mathcal {L}}{\phi (t)} ]

This transformation yields:

[ {\mathcal {L}}{f(t)}\sum {i=0}^{n}a{i}s^{i} - \sum _{i=1}^{n}\sum {j=1}^{i}a{i}s^{i-j}f^{(j-1)}(0) = {\mathcal {L}}{\phi (t)} ]

Solving for ({\mathcal {L}}{f(t)}) and substituting ( f^{(i)}(0) ) with ( c_{i} ), we obtain:

[ {\mathcal {L}}{f(t)} = \frac{{\mathcal {L}}{\phi (t)} + \sum {i=1}^{n}\sum {j=1}^{i}a{i}s^{i-j}c{j-1}}{\sum {i=0}^{n}a{i}s^{i}} ]

The solution for ( f(t) ) is then obtained by applying the inverse Laplace transform to ({\mathcal {L}}{f(t)}).

It is worth noting that if the initial conditions are all zero, i.e.,

[ f^{(i)}(0) = c_{i} = 0 \quad \forall i \in {0,1,2,…\ n} ]

then the formula simplifies to:

[ f(t) = {\mathcal {L}}^{-1}\left{\frac{{\mathcal {L}}{\phi (t)}}{\sum {i=0}^{n}a{i}s^{i}}\right} ]

An Example

To illustrate the practical application of this method, let’s solve the following differential equation:

[ f’’(t) + 4f(t) = \sin(2t) ]

with initial conditions ( f(0) = 0 ) and ( f’(0) = 0 ).

First, we identify the forcing function:

[ \phi (t) = \sin(2t) ]

Taking the Laplace transform of (\phi (t)):

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

The differential equation in the s-domain becomes:

[ s^{2}{\mathcal {L}}{f(t)} - sf(0) - f’(0) + 4{\mathcal {L}}{f(t)} = {\mathcal {L}}{\phi (t)} ]

Substituting the initial conditions and simplifying:

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

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

Finally, applying the inverse Laplace transform:

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