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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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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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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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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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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

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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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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

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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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Solving second order linear differential equations using the Laplace transform

Plot of  y(t)  vs  t:

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30

20

10

0

10

20

30

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