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Solving second order linear differential equations using the Laplace transform
Solving second order linear differential equations
using the Laplace transform
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Problem 1
d2y
dt2 + y = cos(t)
y(0) = 0, y(0) = 0
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Take Laplace transforms of both sides:
L
d2y
dt2
+ L[y] = L[cos(t)]
s2L[y] − sy(0) − y(0) + L[y] = s
s2
+ 1
s2L[y] + L[y] = s
s2 + 1
L[y] = s
(s2
+ 1)2
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Need to find
L−1 s
(s2 + 1)2
From a table of Laplace transforms,
y(t) = t sin(ωt) ⇐⇒ L[y] = 2ωs
(s2
+ ω2
)2
This is almost our formula with ω = 1.
L−1 s
(s2 + 1)2 = 1
2L−1
2s
(s2 + 1)2 = 1
2t sin(t)
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
So, the solution is y(t) = 12t sin(t).
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20
10
10
20
30
S l i d d li diff i l i i h L l f
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Problem 2
d2y
dt2 + y = cos(t)(1 − u10π(t))
y(0) = 0, y(0) = 0
where u10π(t) is the heavyside function
u10π(t) =
0, t
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Forcing function “turns off” at time t = 10π
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1.0
0.5
0.5
1.0
Sol ing second order linear differential eq ations sing the Laplace transform
http://find/
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Take Laplace transform of both sides:
L
d2y
dt2
+ L[y] = L[cos(t)(1 − u10π(t))]
s2
L[y] − sy(0) − y
(0) + L[y] = L[cos(t)(1 − u10π(t))]s2L[y] + L[y] = L[cos(t)(1 − u10π(t))]
Focusing on right hand side,
L[cos(t)(1 − u10π(t))] = L[cos(t)] − L[u10π(t)cos(t)]
L[cos(t)] = s
s2 + 1
Solving second order linear differential equations using the Laplace transform
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Laplace transforms of forms containing heavyside functions:
L[ua(t)f (t − a)] = e−asL[f ]
Notice that cos(t) = cos(t − 10π), so
L[u10π(t)cos(t)] = L[u10π(t)cos(t − 10π)] = e−10πt
s
s2 + 1
So the Laplace transform of our equation is
L[y](s2 + 1) = s
s2 + 1 − e−10πs
s
s2 + 1
=⇒ L[y] = s(s2 + 1)2
− e−10πs s(s2 + 1)2
Solving second order linear differential equations using the Laplace transform
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Now take inverse Laplace transform of both sides
y(t) = L−1
s
(s2 + 1)2
− L−1
e−10πs
s
(s2 + 1)2
The first piece we found in problem 1
L−1
s(s2 + 1)2
= 1
2t sin(t)
For the second piece, we again apply the formula
L[ua(t)f (t − a)] = e−as
L[f ]
but this time in reverse.
Solving second order linear differential equations using the Laplace transform
http://find/
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Using “L[ua(t)f (t − a)] = e−asL[f ]”, we have a = 10π, and
L[f ] = s(s2 + 1)2
=⇒ f (t) = 12t sin(t)
=⇒ f (t − 10π) = 1
2(t − 10π) sin(t − 10π)
=⇒ L−1e−10πs
s
(s2 + 1)2
= 1
2u10π(t)(t − 10π) sin(t)
Putting all of this together,
y(t) = L−1 s
(s2 + 1)2
− L−1e−10πs
s
(s2 + 1)2
= 1
2 sin(t) [t − u10π(t)(t − 10π)]
Solving second order linear differential equations using the Laplace transform
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8/13/2019 Laplace Transform No Pause Powerpoint for Process Design
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Solving second order linear differential equations using the Laplace transform
Plot of y(t) vs t:
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20
10
0
10
20
30
http://find/
