ch 6.4: differential equations with discontinuous forcing functions in this section, we focus on...
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![Page 1: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/1.jpg)
Ch 6.4: Differential Equations with Discontinuous Forcing Functions
In this section, we focus on examples of nonhomogeneous initial value problems in which the forcing function is discontinuous.
00 0,0),( yyyytgcyybya
![Page 2: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/2.jpg)
Example 1: Initial Value Problem (1 of 12)
Find the solution to the initial value problem
Such an initial value problem might model the response of a damped oscillator subject to g(t), or current in a circuit for a unit voltage pulse.
20 and 50,0
205,1)()()(
where
0)0(,0)0(),(22
205 tt
ttututg
yytgyyy
![Page 3: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/3.jpg)
Assume the conditions of Corollary 6.2.2 are met. Then
or
Letting Y(s) = L{y},
Substituting in the initial conditions, we obtain
Thus
)}({)}({}{2}{}{2 205 tuLtuLyLyLyL
s
eeyLyysLysyyLs
ss 2052 }{2)0(}{)0(2)0(2}{2
seeyyssYss ss 2052 )0(2)0(12)(22
Example 1: Laplace Transform (2 of 12)
seesYss ss 2052 )(22
22
)(2
205
sss
eesY
ss
0)0(,0)0(),()(22 205 yytutuyyy
![Page 4: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/4.jpg)
We have
where
If we let h(t) = L-1{H(s)}, then
by Theorem 6.3.1.
)20()()5()()( 205 thtuthtuty
Example 1: Factoring Y(s) (3 of 12)
)(
22)( 205
2
205
sHeesss
eesY ss
ss
22
1)(
2
ssssH
![Page 5: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/5.jpg)
Thus we examine H(s), as follows.
This partial fraction expansion yields the equations
Thus
Example 1: Partial Fractions (4 of 12)
2222
1)(
22
ss
CBs
s
A
ssssH
2/1,1,2/1
12)()2( 2
CBA
AsCAsBA
22
2/12/1)(
2
ss
s
ssH
![Page 6: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/6.jpg)
Completing the square,
Example 1: Completing the Square (5 of 12)
16/154/1
4/14/1
2
12/1
16/154/1
2/1
2
12/1
16/1516/12/
2/1
2
12/1
12/
2/1
2
12/122
2/12/1)(
2
2
2
2
2
s
s
s
s
s
s
ss
s
s
ss
s
s
ss
s
ssH
![Page 7: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/7.jpg)
Thus
and hence
For h(t) as given above, and recalling our previous results, the solution to the initial value problem is then
Example 1: Solution (6 of 12)
16/154/1
4/15
152
1
16/154/1
4/1
2
12/1
16/154/1
4/14/1
2
12/1)(
22
2
ss
s
s
s
s
ssH
tetesHLth tt
4
15sin
152
1
4
15cos
2
1
2
1)}({)( 4/4/1
)20()()5()()( 205 thtuthtut
![Page 8: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/8.jpg)
Thus the solution to the initial value problem is
The graph of this solution is given below.
Example 1: Solution Graph (7 of 12)
415sin152
1415cos
2
1
2
1)(
where),20()()5()()(
4/4/
205
teteth
thtuthtut
tt
![Page 9: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/9.jpg)
Find the solution to the initial value problem
The graph of forcing function
g(t) is given on right, and is
known as ramp loading.
10,1
10555
50,0
5
10)(
5
5)()(
where
0)0(,0)0(),(4
105
t
tt
tt
tut
tutg
yytgyy
Example 2: Initial Value Problem (1 of 12)
![Page 10: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/10.jpg)
Assume that this ODE has a solution y = (t) and that '(t) and ''(t) satisfy the conditions of Corollary 6.2.2. Then
or
Letting Y(s) = L{y}, and substituting in initial conditions,
Thus
5}10)({5}5)({}{4}{ 105 ttuLttuLyLyL
2
1052
5}{4)0()0(}{
s
eeyLysyyLs
ss
Example 2: Laplace Transform (2 of 12)
21052 5)(4 seesYs ss
45
)(22
105
ss
eesY
ss
0)0(,0)0(,5
10)(
5
5)(4 105
yy
ttu
ttuyy
![Page 11: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/11.jpg)
We have
where
If we let h(t) = L-1{H(s)}, then
by Theorem 6.3.1.
)10()()5()(5
1)( 105 thtuthtuty
Example 2: Factoring Y(s) (3 of 12)
)(
545)(
105
22
105
sHee
ss
eesY
ssss
4
1)(
22
sssH
![Page 12: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/12.jpg)
Thus we examine H(s), as follows.
This partial fraction expansion yields the equations
Thus
Example 2: Partial Fractions (4 of 12)
44
1)(
2222
s
DCs
s
B
s
A
sssH
4/1,0,4/1,0
144)()( 23
DCBA
BAssDBsCA
4
4/14/1)(
22
sssH
![Page 13: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/13.jpg)
Thus
and hence
For h(t) as given above, and recalling our previous results, the solution to the initial value problem is then
Example 2: Solution (5 of 12)
ttsHLth 2sin8
1
4
1)}({)( 1
4
2
8
11
4
14
4/14/1)(
22
22
ss
sssH
)10()()5()(5
1)( 105 thtuthtuty
![Page 14: Ch 6.4: Differential Equations with Discontinuous Forcing Functions In this section, we focus on examples of nonhomogeneous initial value problems in which](https://reader036.vdocuments.mx/reader036/viewer/2022082701/551a9a59550346e0158b55c3/html5/thumbnails/14.jpg)
Thus the solution to the initial value problem is
The graph of this solution is given below.
Example 2: Graph of Solution (6 of 12)
ttth
thtuthtut
2sin8
1
4
1)(
where,)10()()5()(5
1)( 105