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Chapter 7. Laplace Transforms. Section 7.4 Inverse Laplace Transform.
Definition 1. Given a function F (s), if there is a function f (t) that is continuous on [0, ∞) and satisfies L{f }(s) = F (s), then we say that f (t) is the inverse Laplace transform of F (s) and employ the notation f (t) = L−^1 {F }(t).
Table of inverse Laplace transform
F (s) f (t) = L−^1 {F }(t) 1 s ,^ s >^0 1 s − a,^ s > a^ e
at (n − 1)! sn^ ,^ s >^0 t
n− (^1) , n = 1, 2 , ... b s^2 + b^2
, s > 0 sin bt s s^2 + b^2 ,^ s >^0 cos^ bt (n − 1)! (s − a)n^ ,^ s > a^ e
attn− (^1) , n = 1, 2 , ...
b (s − a)^2 + b^2
, s > a eat^ sin bt s − a (s − a)^2 + b^2
, s > a eat^ cos bt
Example 1. Determine the inverse Laplace transform of the given function. (a) F (s) = (^) s^23. SOLUTION. L−^1
s^3
= L−^1
s^3
= t^2
(b) F (s) = (^) s (^22) +. SOLUTION. L−^1
s^2 +
= L−^1
s^2 +2^2
= sin 2t.
(c) F (s) = (^) s (^2) +2s+1s+.
SOLUTION. L−^1
{ (^) s+ s^2 +2s+
= L−^1
s+ (s+1)^2 +
= L−^1
s+ (s+1)^2 +3^2
= e−t^ cos 3t.
Theorem 1. (linearity of the inverse transform) Assume that L−^1 {F }, L−^1 {F 1 }, and L−^1 {F 2 } exist and are continuous on [0, ∞) and c is any constant. Then
L−^1 {F 1 + F 2 } = L−^1 {F 1 } + L−^1 {F 2 } L−^1 {cF } = cL−^1 {F }.
Example 2. Determine L−^1
3 (2s+5)^3 +^
2 s+ s^2 +4s+13 +^
3 s^2 +4s+
SOLUTION. L−^1
3 (2s+5)^3 +^
2 s+ s^2 +4s+13 +^
3 s^2 +4s+
= L−^1
(2s + 5)^3
+ L−^1
2 s + 16 s^2 + 4s + 13
+ L−^1
s^2 + 4s + 8
= 3L−^1
23 (s + 52 )^3
+ 2L−^1
s + 8 (s + 2)^2 + 9
+ 3L−^1
(s + 2)^2 + 4
L−^1
(s + 52 )^3
+ 2L−^1
(s + 2) + 6 (s + 2)^2 + 3^2
+ 3L−^1
(s + 2)^2 + 2^2
L−^1
(s + 52 )^3
+2L−^1
s + 2 (s + 2)^2 + 3^2
+2L−^1
(s + 2)^2 + 3^2
L−^1
(s + 2)^2 + 2^2
e−^ 25 t t^2 + 2e−^2 t^ cos 3t + 4e−^2 t^ sin 3t +
e−^2 t^ sin 2t =
e−^
(^52) t t^2 + e−^2 t(2 cos 3t + 4 sin 3t +
sin 2t)
Example 3. Determine L−^1 {F }, where (a) F (s) = (^) (s−s1)(^2 −s^26 +2)(s−^47 s+5) , SOLUTION. We begin by finding the partial fraction expantion for F (s). The denominator consists of three linear factors, so the expantion has the form
s^2 − 26 s − 47 (s − 1)(s + 2)(s + 5)
A
s − 1
B
s + 2
C
s + 5
where numbers A, B, and C to be determined.
s^2 − 26 s − 47 (s − 1)(s + 2)(s + 5)
A
s − 1
B
s + 2
C
s + 5
A(s + 2)(s + 5) + B(s − 1)(s + 5) + C(s − 1)(s + 2) (s − 1)(s + 2)(s + 5)
So, we have that
s^2 − 26 s − 47 = A(s + 2)(s + 5) + B(s − 1)(s + 5) + C(s − 1)(s + 2).
We can find A, B, and C plugging s = 1, s = −2, and s = −5 into last equality.
s = 1 : 1 − 26 − 47 = A(1 + 2)(1 + 5), s = −2 : 4 − 26(−2) − 47 = B(− 2 − 1)(−2 + 5), s = −5 : 25 − 26(−5) − 47 = C(− 5 − 1)(−5 + 2).
So, A = −4, B = −1, C = 6. Thus,
s^2 − 26 s − 47 (s − 1)(s + 2)(s + 5)
s − 1
s + 2
s + 5
