If the Laplace transform of a function f(t) is known, then the Laplace transform of its derivative f'(t) can be computed. Actually, one can find the Laplace transform of (any finite order) derivatives of f(t). In this article, we will learn how to find the Laplace transform of derivatives and use it to solve an ordinary differential equation as an application.
Table of Contents
First order derivative
| The Laplace transform of first order derivative is given by the formula: L{f$’$(t)} = s L{f(t)} -f(0). |
Proof:
By the definition of Laplace transforms, we have that
L{f$’$(t)} = $\int_0^\infty e^{-st} f'(t) dt$
Integrating the above by parts, we get that
L{f$’$(t)} = $\Big[ e^{-st} f(t) \Big]_0^\infty$ – $\int_0^\infty (-se^{-st}) f(t) dt$
As e-st f(t)→0 when t→∞, from above it follows that
L{f$’$(t)} = -f(0) +s $\int_0^\infty e^{-st} f(t) dt$
= -f(0) + s L{f(t)}.
So we have proved that L{f$’$(t)} = -f(0) + s L{f(t)} and this is the Laplace transform formula of first order derivative.
Remark: Note that if f(0)=0, then L{f$’$(t)} = s L{f(t)}, that is, the Laplace transform of the derivative of f(t) is equal to the multiplication of the Laplace transform of f(t) by s.
Second order derivative
| The Laplace transform of first order derivative is given by the formula: L{f$^{”}$(t)} = s2 L{f(t)} – sf(0) -f$’$(0). |
Proof:
From above, we have that
L{f$’$(t)} = -f(0) + s L{f(t)} …(∗)
Replacing f(t) by f$’$(t) in this formula, we get that
L{f$^{”}$(t)} = -f$’$(0) + s L{f$’$(t)}
= -f$’$(0) + s (-f(0) + s L{f(t)}) by (∗)
= s2 L{f(t)} – sf(0) -f$’$(0).
Thus, we have shown that L{f$^{”}$(t)} = s2 L{f(t)} – sf(0) -f$’$(0) and this is the Laplace transform formula of second order derivatives.
Related Topics:
Laplace Transform: Definition, Table, Formulas, Properties
Solved problems of Laplace transforms
nth order derivative
| The Laplace transform formula of nth order derivative is given as follows: L{f(n)(t)} = snF(s) – sn-1f(0) – sn-2$f^\prime(0)$ – … -f(n-1)(0), where f(n) denotes the n-th derivative of f(t). |
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