Why is Laplace Transform used for ODEs

Question

This part is taken from differential equations with applications and historical George simmons.

According to the given information , there are another integral transformation.I wonder why is the Laplace transform ($a=0 , b=\infty , K(p,x)=e^{-px}$) begin used for solving ODEs.Are there any advantages or other particular reasons for using this combination ?

Answer 1

The Laplace transform is useful because

1. it transforms an ODE into an algebraic equation in the transformed variable (or a PDE into an ODE), and
2. it includes the initial conditions as part of the algebraic equation.

The usefulness of this LT of course depends on the ability to find the inverse transform of the solution to the algebraic equation. In general, this requires the evaluation of a complex integral. However, for those not versed in complex integration methods, there are tables from which one may simply write down the inverse. Thus, the LT provides a simple means of solving an ODE using algebra.

No better way to understand this than to provide a specific example. Let's solve

$$f''(t) + 4 f'(t) + 3 f(t) = t \sin{t}$$ $$f(0) = 0 \quad f'(0) = 1$$

The LT of the LHS of the equation is

$$s^2 F(s) - 1 + 4 s F(s) + 3F(s) $$

The LT of the RHS is

$$\int_0^{\infty} dt \, t \sin{t}\; e^{-s t} = \operatorname{Im}{\int_0^{\infty} dt \, t \, e^{-(s-i) t}} = \operatorname{Im}{\left [\frac1{(s-i)^2}\right ]} = \frac{2 s}{(s^2+1)^2}$$

We need only solve a simple algebraic equation to find $F(s)$:

$$F(s) = \frac{(s^2+1)^2+2 s}{(s^2+1)^2 (s+3)(s+1)} $$

We may then use tables or the residue theorem to find the inverse:

$$f(t) = \frac14 e^{-t} - \frac{47}{100} e^{-3 t} + \frac{1}{50} [(5 t+2) \sin{t} +(11-10 t) \cos{t}]$$

Answer 2

There are in fact many other integral transforms with specific kernels. The truth is that there is virtually no advantage of using the Laplace transform to solve differential equations, in comparison to other methods. Indeed, the method requires that you are able to compute explicitly the Laplace transform of any functions that are involved, which is really too much to hope for. Moreover, the transform has no use in the study of arbitrary perturbations, say to understand whether sufficiently small perturbations don't change the type of stability of an equilibrium point. Overall, in the context of solving differential equations, it is not much to say that the transform is of no use.