Solving differential equations by using the Laplace transform

University of Babylon Lecture: Roaya Mahmood Jaleel
faculty of Engineering Subject: Mathematics
Department of Chemical Engineering Stage : 2nd stage
4
Solving differential equations by using the Laplace transform
the Laplace transforms can be used to solve linear differential equations algebraically.
Theorem: [Laplace transform of derivatives] Suppose f is of exponential
order, and that f is continuous and f ? is piecewise continuous on any interval 0 ? t ? A. Then
L { f ?(t)} = s L {f (t)} ? f (0)
Applying the theorem multiple times yields:
L {f ?(t)} = s2L {f (t)} ? s f (0) ? f ?(0),
L {f ??(t)} = s3L {f (t)} ? s2f (0) ? s f ?(0) ? f ?(0),
L {f(n)(t)} = snL {f (t)} ? sn? 1f (0) ? sn? 2f ?(0) ? …
? s 2f(n ?3)(0) ? s f(n ?2)(0) ? f(n ?1)(0).
University of Babylon Lecture: Roaya Mahmood Jaleel
faculty of Engineering Subject: Mathematics
Department of Chemical Engineering Stage : 2nd stage
5
University of Babylon Lecture: Roaya Mahmood Jaleel
faculty of Engineering Subject: Mathematics
Department of Chemical Engineering Stage : 2nd stage
6
Example: y? ? 6y? + 5y = 0, y(0) = 1, y?(0) = ?3
[Step 1] Transform both sides
L{y? ? 6y? + 5y} = L{0}(s2L{y} ? s y(0) ? y?(0)) ? 6(sL{y} ? y(0)) + 5L{y} = 0
[Step 2] Simplify to find Y(s) = L{y}
(s2L{y} ? s ?(?3)) ? 6(s L{y} ? 1) + 5L{y} = 0
(s2 ? 6 s + 5) L{y} ? s + 9 = 0
(s2 ? 6 s + 5) L{y} = s ? 9
L{y} = (s2 ? 6 s + 5) / s – 9
[Step 3] Find the inverse transform y(t)Use partial fractions to simplify,
L{y} =
s ??9 ??a(s ??5) ??b(s ?1) ??(a ??b)s ??(?5a?b)
Equating the corresponding coefficients:
1 = a + b a = 2
?9 = ?5a – b b = ?1
Hence,
L{y} =
The last expression corresponds to the Laplace transform of
2e t ? e 5t. Therefore, it must be that
y(t) = 2et ? e 5t.
University of Babylon Lecture: Roaya Mahmood Jaleel
faculty of Engineering Subject: Mathematics
Department of Chemical Engineering Stage : 2nd stage
7
Example: y? + 2y = 4t e?2t, y(0) = ?3.
[Step 1] Transform both sides
L{y? + 2y} = L{4t e?2t}
(sL{y} ? y(0)) + 2L{y} = L{4t e ?2t} =4/ (s +2)2
[Step 2] Simplify to find Y(s) = L{y}
(sL{y} ? (?3)) + 2L{y} = 4/ (s +2)2
(s + 2) L{y} + 3 = 4/ (s +2)2
(s + 2) L{y} =
L{y}= =