582 CHAPTER 30. LAPLACE TRANSFORM METHODS

Proof: Let f ,g be two functions having exponential growth. Then for s large enough,

L (a f (t)+bg(t)) ≡∫

∞

0e−ts (a f (t)+bg(t))dt

= a∫

∞

0e−ts f (t)dt +b

∫∞

0e−tsg(t)dt = aL f (s)+bL g(s) ■

The usefulness of this method in solving differential equations, comes from the follow-ing observation.

L(x′ (t)

)=∫

∞

0x′ (t)e−tsdt = x(t)e−st |∞0 +

∫∞

0se−stx(t)dt =−x(0)+ sL x(s) .

In the following table, Γ(p+1) denotes the gamma function

Γ(p+1) =∫

∞

0e−tt pdt

The function uc (t) denotes the step function which equals 1 for t > c and 0 for t < c.

c

1uc(t)

The expression in Formula 20.) is defined as follows∫δ (t− c) f (t)dt = f (c)

It models an impulse and is sometimes called the Dirac delta function. There is no suchfunction but it is called this anyway. In the following, n will be a positive integer andf ∗ g(t) ≡

∫ t0 f (t−u)g(u)du. Also, F (s) will denote L { f (t)} the Laplace transform of

the function t→ f (t).