Mathematical Methods for Physicists: A concise introduction - Site Map

FUNCTIONS DEFINED BY INTEGRALS
Solution: Note that y 0 ˆ dy=dt. We know how to solve this simple di€erential
equation, but as an illustration we now solve it using Laplace transforms. Taking
both sides of the equation we obtain
Now
L‰ y 00 Š‡L‰ yŠ ˆL‰1Š; …L‰ f ŠˆL‰1Š†:
L‰ y 00 ŠˆpL‰ y 0 Šy 0 …0† ˆpfpL‰ yŠy…0†g y 0 …0†
ˆ p 2 L‰ yŠpy…0†y 0 …0†
ˆ p 2 L‰ yŠ
and
L‰1Š ˆ1=p:
The transformed equation then becomes
p 2 Ly ‰ Š‡Ly ‰ Š ˆ 1=p
or
therefore
1
Ly ‰ Š ˆ
p… p 2 ‡ 1† ˆ 1
p
p
p 2 ‡ 1 ;
y ˆ L 1 1
L 1 p
p p 2 :
‡ 1
We ®nd from Eqs. (9.6) and (9.10) that
L 1 1
ˆ 1 and L 1 p
p
p 2 ‡ 1
Thus, the solution of the initial problem is
ˆ cos t:
y ˆ 1 cos t for t 0; y ˆ 0 for t < 0:
Laplace transforms of functions de®ned by integrals
x
If g…x† ˆR 0
f …u†du, and if Lf…x† ‰ Š ˆ F…p†, then Lg…x† ‰ Š ˆ F…p†=p.
Similarly, if L 1 ‰ F… p† Š ˆ f …x†, then L 1 ‰ F… p†=pŠ ˆ g…x†:
x
It is easy to prove this. If g…x† ˆR 0 f …u†du, then g…0† ˆ0; g 0 …x† ˆf …x†. Taking
Laplace transform, we obtain
L‰g 0 …x†Š ˆ L‰ f …x†Š
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