hw 8, math 319, fall 2016ย ยท 2 1 hw 8 1.1 section 6.1 problem 7 find laplace transform of ๐...
TRANSCRIPT
HW 8, Math 319, Fall 2016
Nasser M. Abbasi (Discussion section 44272, Th 4:35PM-5:25 PM)
December 30, 2019
Contents
1 HW 8 21.1 Section 6.1 problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Section 6.1 problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Section 6.1 problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Section 6.1 problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Section 6.2 problem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Section 6.2 problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.7 Section 6.2 problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.8 Section 6.2 problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.9 Section 6.2 problem 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.10 Section 6.2 problem 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.11 Section 6.2 problem 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.12 Section 6.3 problem 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.12.1 Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.12.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.12.3 Part (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.13 Section 6.3 problem 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.14 Section 6.3 problem 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.15 Section 6.3 problem 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.16 Section 6.3 problem 29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.17 Section 6.3 problem 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.18 Section 6.3 problem 31 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.19 Section 6.3 problem 32 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.20 Section 6.3 problem 33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.21 Section 6.4 problem 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.21.1 Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.21.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.21.3 Part (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.21.4 Part(d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1
2
1 HW 8
1.1 Section 6.1 problem 7
Find Laplace Transform of ๐ (๐ก) = cosh (๐๐ก)
solution Since cosh (๐๐ก) = ๐๐๐ก+๐โ๐๐ก
2 then
โ cosh (๐๐ก) = 12โ๏ฟฝ๐๐๐ก + ๐โ๐๐ก๏ฟฝ
=12๏ฟฝโ ๐๐๐ก +โ๐โ๐๐ก๏ฟฝ
But
โ๐๐๐ก =1
๐ โ ๐
For ๐ > ๐ and
โ๐๐๐ก =1
๐ โ ๐
For ๐ < ๐. Hence
โ cosh (๐๐ก) = 12 ๏ฟฝ
1๐ โ ๐
+1
๐ + ๐๏ฟฝ
=๐ 2
๐ 2 โ ๐2
For ๐ > |๐|
1.2 Section 6.1 problem 8
Find Laplace Transform of ๐ (๐ก) = sinh (๐๐ก)
solution Since sinh (๐๐ก) = ๐๐๐กโ๐โ๐๐ก
2 then
โ sinh (๐๐ก) = 12โ๏ฟฝ๐๐๐ก โ ๐โ๐๐ก๏ฟฝ
=12๏ฟฝโ ๐๐๐ก โโ ๐โ๐๐ก๏ฟฝ
But, as we found in the last problem
โ๐๐๐ก =1
๐ โ ๐๐ > ๐
And
โ๐โ๐๐ก =1
๐ + ๐๐ < ๐
3
Therefore
โ sinh (๐๐ก) = 12 ๏ฟฝ
1๐ โ ๐
โ1
๐ + ๐๏ฟฝ๐ > ๐; ๐ < ๐
=๐
๐ 2 โ ๐2๐ > |๐|
1.3 Section 6.1 problem 9
Find Laplace Transform of ๐ (๐ก) = ๐๐๐ก cosh (๐๐ก)
solution Using the property that
๐๐๐ก๐ (๐ก)โบ ๐น (๐ โ ๐)
Where ๐ (๐ก) = cosh (๐๐ก) now. We already found above that cosh (๐๐ก)โบ ๐ ๐ 2โ๐2 , for ๐ > |๐|. In other words,
๐น (๐ ) = ๐ ๐ 2โ๐2 , therefore
๐๐๐ก cosh (๐๐ก)โบ(๐ โ ๐)
(๐ โ ๐)2 โ ๐2๐ โ ๐ > |๐|
1.4 Section 6.1 problem 10
Find Laplace Transform of ๐ (๐ก) = ๐๐๐ก sinh (๐๐ก)
solution Using the property that
๐๐๐ก๐ (๐ก)โบ ๐น (๐ โ ๐)
Where ๐ (๐ก) = sinh (๐๐ก) now. We already found above that sinh (๐๐ก)โบ ๐๐ 2โ๐2 , for ๐ > |๐|. In other words,
๐น (๐ ) = ๐๐ 2โ๐2 , therefore
๐๐๐ก sinh (๐๐ก)โบ ๐(๐ โ ๐)2 โ ๐2
๐ โ ๐ > |๐|
1.5 Section 6.2 problem 17
Use Laplace transform to solve ๐ฆ(4) โ4๐ฆโฒโฒโฒ +6๐ฆโฒโฒ โ4๐ฆโฒ +๐ฆ = 0 for ๐ฆ (0) = 0, ๐ฆโฒ (0) = 1, ๐ฆโฒโฒ (0) = 0, ๐ฆโฒโฒโฒ (0) = 1
Solution Taking Laplace transform of the ODE gives
โ๏ฟฝ๐ฆ(4)๏ฟฝ โ 4โ๏ฟฝ๐ฆโฒโฒโฒ๏ฟฝ + 6โ๏ฟฝ๐ฆโฒโฒ๏ฟฝ โ 4โ๏ฟฝ๐ฆโฒ๏ฟฝ +โ๏ฟฝ๐ฆ๏ฟฝ = 0 (1)
Let โ๏ฟฝ๐ฆ๏ฟฝ = ๐ (๐ ) then
โ๏ฟฝ๐ฆ(4)๏ฟฝ = ๐ 4๐ (๐ ) โ ๐ 3๐ฆ (0) โ ๐ 2๐ฆโฒ (0) โ ๐ ๐ฆโฒโฒ (0) โ ๐ฆโฒโฒโฒ (0)
= ๐ 4๐ (๐ ) โ ๐ 3 (0) โ ๐ 2 (1) โ ๐ (0) โ 1= ๐ 4๐ (๐ ) โ ๐ 2 โ 1
And
โ๏ฟฝ๐ฆโฒโฒโฒ๏ฟฝ = ๐ 3๐ (๐ ) โ ๐ 2๐ฆ (0) โ ๐ ๐ฆโฒ (0) โ ๐ฆโฒโฒ (0)
= ๐ 3๐ (๐ ) โ ๐ 2 (0) โ ๐ (1) โ 0= ๐ 3๐ (๐ ) โ ๐
4
And
โ๏ฟฝ๐ฆโฒโฒ๏ฟฝ = ๐ 2๐ (๐ ) โ ๐ ๐ฆ (0) โ ๐ฆโฒ (0)
= ๐ 2๐ (๐ ) โ ๐ (0) โ 1= ๐ 2๐ (๐ ) โ 1
And
โ๏ฟฝ๐ฆโฒ๏ฟฝ = ๐ ๐ (๐ ) โ ๐ฆ (0)
= ๐ ๐ (๐ )
Hence (1) becomes
๏ฟฝ๐ 4๐ (๐ ) โ ๐ 2 โ 1๏ฟฝ โ 4 ๏ฟฝ๐ 3๐ (๐ ) โ ๐ ๏ฟฝ + 6 ๏ฟฝ๐ 2๐ (๐ ) โ 1๏ฟฝ โ 4 (๐ ๐ (๐ )) + ๐ (๐ ) = 0
๐ (๐ ) ๏ฟฝ๐ 4 โ 4๐ 3 + 6๐ 2 โ 4๐ + 1๏ฟฝ โ ๐ 2 โ 1 + 4๐ โ 6 = 0
Therefore
๐ (๐ ) =๐ 2 โ 4๐ + 7
๐ 4 โ 4๐ 3 + 6๐ 2 โ 4๐ + 1
=๐ 2 โ 4๐ + 7(๐ โ 1)4
=๐ 2
(๐ โ 1)4โ
4๐ (๐ โ 1)4
+7
(๐ โ 1)4(2)
But
๐ 2
(๐ โ 1)4=(๐ โ 1)2 โ 1 + 2๐
(๐ โ 1)4
=(๐ โ 1)2
(๐ โ 1)4โ
1(๐ โ 1)4
+ 2(๐ โ 1) + 1(๐ โ 1)4
=1
(๐ โ 1)2โ
1(๐ โ 1)4
+ 2(๐ โ 1)(๐ โ 1)4
+ 21
(๐ โ 1)4
=1
(๐ โ 1)2โ
1(๐ โ 1)4
+ 21
(๐ โ 1)3+ 2
1(๐ โ 1)4
=1
(๐ โ 1)2+
2(๐ โ 1)3
+1
(๐ โ 1)4
And4๐
(๐ โ 1)4= 4
(๐ โ 1) + 1(๐ โ 1)4
= 4(๐ โ 1)(๐ โ 1)4
+ 41
(๐ โ 1)4
=4
(๐ โ 1)3+
4(๐ โ 1)4
Therefore (2) becomes
๐ (๐ ) = ๏ฟฝ1
(๐ โ 1)2+
2(๐ โ 1)3
+1
(๐ โ 1)4๏ฟฝ โ ๏ฟฝ
4(๐ โ 1)3
+4
(๐ โ 1)4๏ฟฝ +
7(๐ โ 1)4
=1
(๐ โ 1)2โ
2(๐ โ 1)3
+4
(๐ โ 1)4(3)
5
Now using property the shift property of ๐น (๐ ) together with1๐ 2โบ ๐ก
1๐ 3โบ
๐ก2
21๐ 4โบ
๐ก3
6Therefore
1(๐ โ 1)2
โบ๐๐ก๐ก
1(๐ โ 1)3
โบ๐๐ก๐ก2
21
(๐ โ 1)4โบ๐๐ก
๐ก3
6
And (3) becomes
1(๐ โ 1)2
โ2
(๐ โ 1)3+
4(๐ โ 1)4
โบ๐๐ก๐ก โ 2 ๏ฟฝ๐๐ก๐ก2
2 ๏ฟฝ+ 4 ๏ฟฝ๐๐ก
๐ก3
6 ๏ฟฝ
= ๐๐ก๐ก โ ๐๐ก๐ก2 +23๐๐ก๐ก3
Hence
๐ฆ (๐ก) = ๐๐ก ๏ฟฝ๐ก โ ๐ก2 +23๐ก3๏ฟฝ
1.6 Section 6.2 problem 18
Use Laplace transform to solve ๐ฆ(4) โ ๐ฆ = 0 for ๐ฆ (0) = 1, ๐ฆโฒ (0) = 0, ๐ฆโฒโฒ (0) = 1, ๐ฆโฒโฒโฒ (0) = 0
Solution Taking Laplace transform of the ODE gives
โ๏ฟฝ๐ฆ(4)๏ฟฝ โโ๏ฟฝ๐ฆ๏ฟฝ = 0 (1)
Let โ๏ฟฝ๐ฆ๏ฟฝ = ๐ (๐ ) then
โ๏ฟฝ๐ฆ(4)๏ฟฝ = ๐ 4๐ (๐ ) โ ๐ 3๐ฆ (0) โ ๐ 2๐ฆโฒ (0) โ ๐ ๐ฆโฒโฒ (0) โ ๐ฆโฒโฒโฒ (0)
= ๐ 4๐ (๐ ) โ ๐ 3 (1) โ ๐ 2 (0) โ ๐ (1) โ 0= ๐ 4๐ (๐ ) โ ๐ 3 โ ๐
Hence (1) becomes
๐ 4๐ (๐ ) โ ๐ 3 โ ๐ โ ๐ (๐ ) = 0
6
Solving for ๐ (๐ ) gives
๐ (๐ ) =๐ 3 + ๐ ๐ 4 โ 1
=๐ ๏ฟฝ๐ 2 + 1๏ฟฝ๐ 4 โ 1
=๐ ๏ฟฝ๐ 2 + 1๏ฟฝ
๏ฟฝ๐ 2 โ 1๏ฟฝ ๏ฟฝ๐ 2 + 1๏ฟฝ
=๐
๐ 2 โ 1But, Hence above becomes, where ๐ = 1
๐ ๐ 2 โ 1
โบ cosh (๐ก)
Hence
๐ฆ (๐ก) = cosh (๐๐ก)
1.7 Section 6.2 problem 19
Use Laplace transform to solve ๐ฆ(4) โ 4๐ฆ = 0 for ๐ฆ (0) = 1, ๐ฆโฒ (0) = 0, ๐ฆโฒโฒ (0) = โ2, ๐ฆโฒโฒโฒ (0) = 0
Solution Taking Laplace transform of the ODE gives
โ๏ฟฝ๐ฆ(4)๏ฟฝ โ 4โ๏ฟฝ๐ฆ๏ฟฝ = 0 (1)
Let โ๏ฟฝ๐ฆ๏ฟฝ = ๐ (๐ ) then
โ๏ฟฝ๐ฆ(4)๏ฟฝ = ๐ 4๐ (๐ ) โ ๐ 3๐ฆ (0) โ ๐ 2๐ฆโฒ (0) โ ๐ ๐ฆโฒโฒ (0) โ ๐ฆโฒโฒโฒ (0)
= ๐ 4๐ (๐ ) โ ๐ 3 (1) โ ๐ 2 (0) โ ๐ (โ2) โ 0= ๐ 4๐ (๐ ) โ ๐ 3 + 2๐
Hence (1) becomes
๐ 4๐ (๐ ) โ ๐ 3 + 2๐ โ 4๐ (๐ ) = 0
Solving for ๐ (๐ ) gives
๐ (๐ ) =๐ 3 โ 2๐ ๐ 4 โ 4
=๐ 3 โ 2๐
๏ฟฝ๐ 2 โ 2๏ฟฝ ๏ฟฝ๐ 2 + 2๏ฟฝ
=๐ ๏ฟฝ๐ 2 โ 2๏ฟฝ
๏ฟฝ๐ 2 โ 2๏ฟฝ ๏ฟฝ๐ 2 + 2๏ฟฝ
=๐
๏ฟฝ๐ 2 + 2๏ฟฝ
Using cos (๐๐ก)โบ ๐ ๐ 2+๐2 , the above becomes, where ๐ = โ2
๐ ๏ฟฝ๐ 2 + 2๏ฟฝ
โบ cos ๏ฟฝโ2๐ก๏ฟฝ
7
Hence
๐ฆ (๐ก) = cos ๏ฟฝโ2๐ก๏ฟฝ
1.8 Section 6.2 problem 20
Use Laplace transform to solve ๐ฆโฒโฒ + ๐2๐ฆ = cos 2๐ก; ๐2 โ 4; ๐ฆ (0) = 1, ๐ฆโฒ (0) = 0
Solution Let ๐ (๐ ) = โ๏ฟฝ๐ฆ (๐ก)๏ฟฝ. Taking Laplace transform of the ODE, and using cos (๐๐ก)โบ ๐ ๐ 2+๐2 gives
๐ 2๐ (๐ ) โ ๐ ๐ฆ (0) โ ๐ฆโฒ (0) + ๐2๐ (๐ ) =๐
๐ 2 + 4(1)
Applying initial conditions
๐ 2๐ (๐ ) โ ๐ + ๐2๐ (๐ ) =๐
๐ 2 + 4Solving for ๐ (๐ )
๐ (๐ ) ๏ฟฝ๐ 2 + ๐2๏ฟฝ โ ๐ =๐
๐ 2 + 4๐ (๐ ) =
๐ ๏ฟฝ๐ 2 + 4๏ฟฝ ๏ฟฝ๐ 2 + ๐2๏ฟฝ
+๐
๏ฟฝ๐ 2 + ๐2๏ฟฝ(2)
But๐
๏ฟฝ๐ 2 + 4๏ฟฝ ๏ฟฝ๐ 2 + ๐2๏ฟฝ=๐ด๐ + ๐ต๏ฟฝ๐ 2 + 4๏ฟฝ
+๐ถ๐ + ๐ท๏ฟฝ๐ 2 + ๐2๏ฟฝ
๐ = (๐ด๐ + ๐ต) ๏ฟฝ๐ 2 + ๐2๏ฟฝ + (๐ถ๐ + ๐ท) ๏ฟฝ๐ 2 + 4๏ฟฝ
๐ = 4๐ท + ๐ด๐ 3 + ๐ต๐ 2 + ๐ถ๐ 3 + ๐ต๐2 + ๐ 2๐ท + 4๐ถ๐ + ๐ด๐ ๐2
๐ = ๏ฟฝ4๐ท + ๐ต๐2๏ฟฝ + ๐ ๏ฟฝ4๐ถ + ๐ด๐2๏ฟฝ + ๐ 2 (๐ต + ๐ท) + ๐ 3 (๐ด + ๐ถ)
Hence
4๐ท + ๐ต๐2 = 04๐ถ + ๐ด๐2 = 1
๐ต + ๐ท = 0๐ด + ๐ถ = 0
Equation (2,4) gives ๐ด = 1๐2โ4 , ๐ถ =
14โ๐2 and (1,3) gives ๐ต = 0,๐ท = 0. Hence
๐ ๏ฟฝ๐ 2 + 4๏ฟฝ ๏ฟฝ๐ 2 + ๐2๏ฟฝ
= ๏ฟฝ1
๐2 โ 4๏ฟฝ๐
๏ฟฝ๐ 2 + 4๏ฟฝ+ ๏ฟฝ
14 โ ๐2 ๏ฟฝ
๐ ๏ฟฝ๐ 2 + ๐2๏ฟฝ
Therefore (2) becomes
๐ (๐ ) = ๏ฟฝ1
๐2 โ 4๏ฟฝ๐
๏ฟฝ๐ 2 + 4๏ฟฝ+ ๏ฟฝ
14 โ ๐2 ๏ฟฝ
๐ ๏ฟฝ๐ 2 + ๐2๏ฟฝ
+๐
๏ฟฝ๐ 2 + ๐2๏ฟฝ
= ๏ฟฝ1
๐2 โ 4๏ฟฝ๐
๏ฟฝ๐ 2 + 4๏ฟฝ+ ๏ฟฝ
5 โ ๐2
4 โ ๐2 ๏ฟฝ๐
๏ฟฝ๐ 2 + ๐2๏ฟฝ
8
Using cos (๐๐ก)โบ ๐ ๐ 2+๐2 , the above becomes
๏ฟฝ1
๐2 โ 4๏ฟฝ๐
๏ฟฝ๐ 2 + 4๏ฟฝ+ ๏ฟฝ
5 โ ๐2
4 โ ๐2 ๏ฟฝ๐
๏ฟฝ๐ 2 + ๐2๏ฟฝโบ ๏ฟฝ
1๐2 โ 4๏ฟฝ
cos (2๐ก) + ๏ฟฝ5 โ ๐2
4 โ ๐2 ๏ฟฝ cos (๐๐ก)
= ๏ฟฝ1
๐2 โ 4๏ฟฝcos (2๐ก) + ๏ฟฝ
๐2 โ 5๐2 โ 4๏ฟฝ
cos (๐๐ก)
Hence
๐ฆ (๐ก) = ๏ฟฝ1
๐2 โ 4๏ฟฝcos (2๐ก) + ๏ฟฝ
๐2 โ 5๐2 โ 4๏ฟฝ
cos (๐๐ก)
=๏ฟฝ๐2 โ 5๏ฟฝ cos (๐๐ก) + cos (2๐ก)
๐2 โ 4
1.9 Section 6.2 problem 21
Use Laplace transform to solve ๐ฆโฒโฒ โ 2๐ฆโฒ + 2๐ฆ = cos ๐ก; ๐ฆ (0) = 1, ๐ฆโฒ (0) = 0
Solution Let ๐ (๐ ) = โ๏ฟฝ๐ฆ (๐ก)๏ฟฝ. Taking Laplace transform of the ODE, and using cos (๐๐ก)โบ ๐ ๐ 2+๐2 gives
๏ฟฝ๐ 2๐ (๐ ) โ ๐ ๐ฆ (0) โ ๐ฆโฒ (0)๏ฟฝ โ 2 ๏ฟฝ๐ ๐ (๐ ) โ ๐ฆ (0)๏ฟฝ + 2๐ (๐ ) =๐
๐ 2 + 1(1)
Applying initial conditions
๐ 2๐ (๐ ) โ ๐ โ 2 (๐ ๐ (๐ ) โ 1) + 2๐ (๐ ) =๐
๐ 2 + 1Solving for ๐ (๐ )
๐ 2๐ (๐ ) โ ๐ โ 2๐ ๐ (๐ ) + 2 + 2๐ (๐ ) =๐
๐ 2 + 1๐ (๐ ) ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ โ ๐ + 2 =
๐ ๐ 2 + 1
๐ (๐ ) =๐
๏ฟฝ๐ 2 + 1๏ฟฝ ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ+
๐ ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ
โ2
๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ(2)
But๐
๏ฟฝ๐ 2 + 1๏ฟฝ ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ=๐ด๐ + ๐ต๏ฟฝ๐ 2 + 1๏ฟฝ
+๐ถ๐ + ๐ท
๐ 2 โ 2๐ + 2
๐ = (๐ด๐ + ๐ต) ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ + (๐ถ๐ + ๐ท) ๏ฟฝ๐ 2 + 1๏ฟฝ
๐ = 2๐ต + ๐ท โ 2๐ด๐ 2 + ๐ด๐ 3 + ๐ต๐ 2 + ๐ถ๐ 3 + ๐ 2๐ท + 2๐ด๐ โ 2๐ต๐ + ๐ถ๐ ๐ = (2๐ต + ๐ท) + ๐ (2๐ด โ 2๐ต + ๐ถ) + ๐ 2 (โ2๐ด + ๐ต + ๐ท) + ๐ 3 (๐ด + ๐ถ)
Hence
2๐ต + ๐ท = 02๐ด โ 2๐ต + ๐ถ = 1โ2๐ด + ๐ต + ๐ท = 0
๐ด + ๐ถ = 0
9
Solving gives ๐ด = 15 , ๐ต = โ
25 , ๐ถ = โ
15 , ๐ท = 4
5 , hence
๐ ๏ฟฝ๐ 2 + 1๏ฟฝ ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ
=15
๐ โ 2๏ฟฝ๐ 2 + 1๏ฟฝ
โ15
๐ โ 4๐ 2 โ 2๐ + 2
=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15
๐ ๐ 2 โ 2๐ + 2
+45
1๐ 2 โ 2๐ + 2
(3)
Completing the squares for
๐ 2 โ 2๐ + 2 = ๐ (๐ + ๐)2 + ๐
= ๐ ๏ฟฝ๐ 2 + ๐2 + 2๐๐ ๏ฟฝ + ๐
= ๐๐ 2 + ๐๐2 + 2๐๐๐ + ๐
Hence ๐ = 1, 2๐๐ = โ2, ๏ฟฝ๐๐2 + ๐๏ฟฝ = 2, hence ๐ = โ1, ๐ = 1, hence
๐ 2 โ 2๐ + 2 = (๐ โ 1)2 + 1
Hence (3) becomes๐
๏ฟฝ๐ 2 + 1๏ฟฝ ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15
๐ (๐ โ 1)2 + 1
+45
1(๐ โ 1)2 + 1
=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15(๐ โ 1) + 1(๐ โ 1)2 + 1
+45
1(๐ โ 1)2 + 1
=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
โ15
1(๐ โ 1)2 + 1
+45
1(๐ โ 1)2 + 1
=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
+35
1(๐ โ 1)2 + 1
Therefore (2) becomes
๐ (๐ ) =15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
+35
1(๐ โ 1)2 + 1
+๐
(๐ โ 1)2 + 1โ
2(๐ โ 1)2 + 1
=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
+35
1(๐ โ 1)2 + 1
+(๐ โ 1) + 1(๐ โ 1)2 + 1
โ2
(๐ โ 1)2 + 1
=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
+35
1(๐ โ 1)2 + 1
+(๐ โ 1)
(๐ โ 1)2 + 1+
1(๐ โ 1)2 + 1
โ2
(๐ โ 1)2 + 1
=15
๐ ๐ 2 + 1
โ25
1๐ 2 + 1
+45
(๐ โ 1)(๐ โ 1)2 + 1
โ25
1(๐ โ 1)2 + 1
Using cos (๐๐ก)โบ ๐ ๐ 2+๐2 , sin (๐๐ก)โบ
๐๐ 2+๐2 and the shift property of Laplace transform, then
15
๐ ๐ 2 + 1
โบ15
cos (๐ก)
25
1๐ 2 + 1
โบ25
sin (๐ก)
45
(๐ โ 1)(๐ โ 1)2 + 1
โบ45๐๐ก cos ๐ก
25
1(๐ โ 1)2 + 1
โบ85๐๐ก sin ๐ก
10
Hence
๐ฆ (๐ก) =15
cos (๐ก) โ 25
sin (๐ก) + 45๐๐ก cos ๐ก โ 2
5๐๐ก sin ๐ก
15๏ฟฝcos ๐ก โ 2 sin ๐ก + 4๐๐ก cos ๐ก โ 2๐๐ก sin ๐ก๏ฟฝ
1.10 Section 6.2 problem 22
Use Laplace transform to solve ๐ฆโฒโฒ โ 2๐ฆโฒ + 2๐ฆ = ๐โ๐ก; ๐ฆ (0) = 0, ๐ฆโฒ (0) = 1
Solution Let ๐ (๐ ) = โ๏ฟฝ๐ฆ (๐ก)๏ฟฝ. Taking Laplace transform of the ODE, and using ๐โ๐ก โบ 1๐ +1 gives
๏ฟฝ๐ 2๐ (๐ ) โ ๐ ๐ฆ (0) โ ๐ฆโฒ (0)๏ฟฝ โ 2 ๏ฟฝ๐ ๐ (๐ ) โ ๐ฆ (0)๏ฟฝ + 2๐ (๐ ) =1
๐ + 1(1)
Applying initial conditions gives
๐ 2๐ (๐ ) โ 1 โ 2๐ ๐ (๐ ) + 2๐ (๐ ) =1
๐ + 1Solving for ๐ (๐ )
๐ (๐ ) ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ โ 1 =1
๐ + 1
๐ (๐ ) =1
(๐ + 1) ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ+
1๐ 2 โ 2๐ + 2
(2)
But1
(๐ + 1) ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ=
๐ด๐ + 1
+๐ต๐ + ๐ถ
๐ 2 โ 2๐ + 2
1 = ๐ด ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ + (๐ต๐ + ๐ถ) (๐ + 1)
1 = 2๐ด + ๐ถ + ๐ด๐ 2 + ๐ต๐ 2 โ 2๐ด๐ + ๐ต๐ + ๐ถ๐ 1 = (2๐ด + ๐ถ) + ๐ (โ2๐ด + ๐ต + ๐ถ) + ๐ 2 (๐ด + ๐ต)
Hence
1 = 2๐ด + ๐ถ0 = โ2๐ด + ๐ต + ๐ถ0 = ๐ด + ๐ต
Solving gives ๐ด = 15 , ๐ต = โ
15 , ๐ถ =
35 , therefore
1(๐ + 1) ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ
=15
1๐ + 1
+โ 15 ๐ +
35
๐ 2 โ 2๐ + 2
=15
1๐ + 1
โ15
๐ ๐ 2 โ 2๐ + 2
+ +35
1๐ 2 โ 2๐ + 2
Completing the square for ๐ 2 โ 2๐ + 2 which was done in last problem, gives (๐ โ 1)2 + 1, hence the
11
above becomes1
(๐ + 1) ๏ฟฝ๐ 2 โ 2๐ + 2๏ฟฝ=15
1๐ + 1
โ15
๐ (๐ โ 1)2 + 1
+ +35
1(๐ โ 1)2 + 1
=15
1๐ + 1
โ15(๐ โ 1) + 1(๐ โ 1)2 + 1
+35
1(๐ โ 1)2 + 1
=15
1๐ + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
โ15
1(๐ โ 1)2 + 1
+35
1(๐ โ 1)2 + 1
=15
1๐ + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
+25
1(๐ โ 1)2 + 1
Therefore (2) becomes
๐ (๐ ) =15
1๐ + 1
โ15
(๐ โ 1)(๐ โ 1)2 + 1
+25
1(๐ โ 1)2 + 1
+1
(๐ โ 1)2 + 1
Using cos (๐๐ก)โบ ๐ ๐ 2+๐2 , sin (๐๐ก)โบ
๐๐ 2+๐2 and the shift property of Laplace transform, then
15
1๐ + 1
โบ15๐โ๐ก
15
(๐ โ 1)(๐ โ 1)2 + 1
โบ15๐๐ก cos ๐ก
25
1(๐ โ 1)2 + 1
โบ25๐๐ก sin ๐ก
1(๐ โ 1)2 + 1
โบ ๐๐ก sin ๐ก
Hence
๐ฆ (๐ก) =15๐โ๐ก โ
15๐๐ก cos ๐ก + 2
5๐๐ก sin ๐ก + ๐๐ก sin ๐ก
=15๏ฟฝ๐โ๐ก โ ๐๐ก cos ๐ก + 7๐๐ก sin ๐ก๏ฟฝ
1.11 Section 6.2 problem 23
Use Laplace transform to solve ๐ฆโฒโฒ + 2๐ฆโฒ + ๐ฆ = 4๐โ๐ก; ๐ฆ (0) = 2, ๐ฆโฒ (0) = โ1
Solution Let ๐ (๐ ) = โ๏ฟฝ๐ฆ (๐ก)๏ฟฝ. Taking Laplace transform of the ODE, and using ๐โ๐ก โบ 1๐ +1 gives
๏ฟฝ๐ 2๐ (๐ ) โ ๐ ๐ฆ (0) โ ๐ฆโฒ (0)๏ฟฝ + 2 ๏ฟฝ๐ ๐ (๐ ) โ ๐ฆ (0)๏ฟฝ + ๐ (๐ ) =4
๐ + 1(1)
Applying initial conditions gives
๏ฟฝ๐ 2๐ (๐ ) โ 2๐ + 1๏ฟฝ + 2 (๐ ๐ (๐ ) โ 2) + ๐ (๐ ) =4
๐ + 1Solving for ๐ (๐ )
๐ (๐ ) ๏ฟฝ๐ 2 + 2๐ + 1๏ฟฝ โ 2๐ + 1 โ 4 =4
๐ + 1
๐ (๐ ) ๏ฟฝ๐ 2 + 2๐ + 1๏ฟฝ =4
๐ + 1+ 2๐ โ 1 + 4
๐ (๐ ) =4
(๐ + 1) ๏ฟฝ๐ 2 + 2๐ + 1๏ฟฝ+
2๐ ๏ฟฝ๐ 2 + 2๐ + 1๏ฟฝ
โ1
๏ฟฝ๐ 2 + 2๐ + 1๏ฟฝ+
4๏ฟฝ๐ 2 + 2๐ + 1๏ฟฝ
12
But ๏ฟฝ๐ 2 + 2๐ + 1๏ฟฝ = (๐ + 1)2, hence
๐ (๐ ) =4
(๐ + 1)3+
2๐ (๐ + 1)2
โ1
(๐ + 1)2+
4(๐ + 1)2
(2)
But2๐
(๐ + 1)2= 2
๐ + 1 โ 1(๐ + 1)2
= 2(๐ + 1)(๐ + 1)2
โ 21
(๐ + 1)2
= 21
๐ + 1โ 2
1(๐ + 1)2
Hence (2) becomes
๐ (๐ ) =4
(๐ + 1)3+ 2
1๐ + 1
โ 21
(๐ + 1)2โ
1(๐ + 1)2
+4
(๐ + 1)2(3)
We now ready to do the inversion. Since 1๐ 3 โบ
๐ก2
2 and 1๐ 2 โบ ๐ก and 1
๐ โบ1 and using the shift property๐๐๐ก๐ (๐ก)โบ ๐น (๐ โ ๐), then using these into (3) gives
4(๐ + 1)3
โบ4๐โ๐ก ๏ฟฝ๐ก2
2 ๏ฟฝ
21
๐ + 1โบ 2๐โ๐ก
21
(๐ + 1)2โบ2๐โ๐ก๐ก
1(๐ + 1)2
โบ๐โ๐ก๐ก
4(๐ + 1)2
โบ4๐โ๐ก๐ก
Now (3) becomes
๐ (๐ )โบ 4๐โ๐ก ๏ฟฝ๐ก2
2 ๏ฟฝ+ 2๐โ๐ก โ 2๐โ๐ก๐ก โ ๐โ๐ก๐ก + 4๐โ๐ก๐ก
= ๐โ๐ก ๏ฟฝ2๐ก2 + 2 โ 2๐ก โ ๐ก + 4๐ก๏ฟฝ
= ๐โ๐ก ๏ฟฝ2๐ก2 + ๐ก + 2๏ฟฝ
1.12 Section 6.3 problem 25
Suppose that ๐น (๐ ) = โ๏ฟฝ๐ (๐ก)๏ฟฝ exists for ๐ > ๐ โฅ 0.
1. Show that if ๐ is positive constant then โ๏ฟฝ๐ (๐๐ก)๏ฟฝ = 1๐๐น ๏ฟฝ
๐ ๐๏ฟฝ for ๐ > ๐๐
2. Show that if ๐ is positive constant then โ โ1 {๐น (๐๐ )} = 1๐๐ ๏ฟฝ
๐ก๐๏ฟฝ
3. Show that if ๐, ๐ are constants with ๐ > 0 then โ โ1 {๐น (๐๐ + ๐)} = 1๐ ๐
โ๐๐ก๐ ๐ ๏ฟฝ ๐ก๐๏ฟฝ
Solution
13
1.12.1 Part (a)
From definition,
โ๏ฟฝ๐ (๐๐ก)๏ฟฝ = ๏ฟฝโ
0๐ (๐๐ก) ๐โ๐ ๐ก๐๐ก
Let ๐๐ก = ๐, then when ๐ก = 0, ๐ = 0 and when ๐ก = โ, ๐ = โ, and ๐ = ๐๐๐๐ก . Hence the above becomes
โ๏ฟฝ๐ (๐๐ก)๏ฟฝ = ๏ฟฝโ
0๐ (๐) ๐โ๐ ๏ฟฝ
๐๐ ๏ฟฝ๐๐๐
=1๐ ๏ฟฝ
โ
0๐ (๐) ๐โ๐๏ฟฝ
๐ ๐ ๏ฟฝ๐๐
We see from above that โ๏ฟฝ๐ (๐๐ก)๏ฟฝ is 1๐๐น ๏ฟฝ
๐ ๐๏ฟฝ .Now we look at the conditions which makes the above
integral converges. Let
๏ฟฝ๐ (๐) ๐โ๐๏ฟฝ๐ ๐ ๏ฟฝ๏ฟฝ โค ๐ ๏ฟฝ๐๐๐ก๐โ๐๏ฟฝ
๐ ๐ ๏ฟฝ๏ฟฝ
Where ๐ is some constant. Then
๏ฟฝโ
0๐ (๐ก) ๐โ๐ก๏ฟฝ
๐ ๐ ๏ฟฝ๐๐ก โค ๐๏ฟฝ
โ
0๐๐๐ก๐โ๐ก๏ฟฝ
๐ ๐ ๏ฟฝ๐๐ก
= ๐๏ฟฝโ
0๐โ๐ก๏ฟฝ
๐ ๐โ๐๏ฟฝ๐๐ก
But โซโ
0๐โ๐ก๏ฟฝ
๐ ๐โ๐๏ฟฝ๐๐ converges if ๐
๐ โ ๐ > 0 or
๐ > ๐๐
Hence this is the condition for โซโ
0๐ (๐ก) ๐โ๐ก๏ฟฝ
๐ ๐ ๏ฟฝ๐๐ก to converge. Which is what we required to show.
1.12.2 Part (b)
From definition
โ๏ฟฝ1๐๐ ๏ฟฝ๐ก๐๏ฟฝ๏ฟฝ =
1๐โ๏ฟฝ๐ ๏ฟฝ
๐ก๐๏ฟฝ๏ฟฝ
=1๐ ๏ฟฝ
โ
0๐ ๏ฟฝ๐ก๐๏ฟฝ ๐โ๐ ๐ก๐๐ก
Let ๐ก๐ = ๐. When ๐ก = 0, ๐ = 0 and when ๐ก = โ, ๐ = โ. ๐๐ก
๐๐ = ๐, hence the above becomes
โ๏ฟฝ1๐๐ ๏ฟฝ๐ก๐๏ฟฝ๏ฟฝ =
1๐ ๏ฟฝ
โ
0๐ (๐) ๐โ๐ (๐๐) (๐๐๐)
= ๏ฟฝโ
0๐ (๐) ๐โ๐(๐ ๐)๐๐
We see from above that โ๏ฟฝ 1๐๐ ๏ฟฝ๐ก๐๏ฟฝ๏ฟฝ is ๐น (๐ ๐). In other words, โ โ1 {๐น (๐๐ )} = 1
๐๐ ๏ฟฝ๐ก๐๏ฟฝ.
14
1.12.3 Part (c)
From definition
โ๏ฟฝ1๐๐โ๐๐ก๐ ๐ ๏ฟฝ
๐ก๐๏ฟฝ๏ฟฝ =
1๐โ๏ฟฝ๐
โ๐๐ก๐ ๐ ๏ฟฝ
๐ก๐๏ฟฝ๏ฟฝ
=1๐ ๏ฟฝ
โ
0๐โ๐๐ก๐ ๐ ๏ฟฝ
๐ก๐๏ฟฝ ๐โ๐ ๐ก๐๐ก
Let ๐ก๐ = ๐, at ๐ก = 0, ๐ = 0 and at ๐ก = โ, ๐ = โ. And ๐๐ก
๐๐ = ๐, hence the above becomes
โ๏ฟฝ1๐๐โ๐๐ก๐ ๐ ๏ฟฝ
๐ก๐๏ฟฝ๏ฟฝ =
1๐ ๏ฟฝ
โ
0๐โ๐(๐๐)
๐ ๐ (๐) ๐โ๐ (๐๐) (๐๐๐)
= ๏ฟฝโ
0๐โ๐๐๐ (๐) ๐โ๐(๐ ๐)๐๐
= ๏ฟฝโ
0๐ (๐) ๐โ๐(๐ ๐+๐)๐๐
We see from the above, that โ๏ฟฝ 1๐ ๐โ๐๐ก๐ ๐ ๏ฟฝ ๐ก๐๏ฟฝ๏ฟฝ = ๐น (๐ ๐ + ๐). Now we look at the conditions which makes
the above integral converges. Let
๏ฟฝ๐ (๐) ๐โ๐ก(๐ ๐+๐)๏ฟฝ โค ๐ ๏ฟฝ๐๐๐ก๐โ๐ก(๐ ๐+๐)๏ฟฝ
Where ๐ is some constant. Then
๏ฟฝโ
0๐ (๐ก) ๐โ๐ก(๐ ๐+๐)๐๐ก โค ๐๏ฟฝ
โ
0๐๐๐ก๐โ๐ก(๐ ๐+๐)๐๐ก
= ๐๏ฟฝโ
0๐โ๐ก(๐ ๐+๐โ๐)๐๐ก
But โซโ
0๐โ๐ก(๐ ๐+๐โ๐)๐๐ก converges if ๐ ๐ + ๐ โ ๐ > 0 or ๐ ๐ > ๐ โ ๐ or ๐ > 1 โ ๐
๐
1.13 Section 6.3 problem 26
Find inverse Laplace transform of ๐น (๐ ) = 2๐+1๐!๐ ๐+1
Solution
We know from tables that๐!๐ ๐+1
โบ ๐ก๐
Hence
2๐+1๐!๐ ๐+1
โบ2๐+1๐ก๐
= 2 (2๐ก)๐
1.14 Section 6.3 problem 27
Find inverse Laplace transform of ๐น (๐ ) = 2๐ +14๐ 2+4๐ +5
Solution
๐น (๐ ) =2๐
4๐ 2 + 4๐ + 5+
14๐ 2 + 4๐ + 5
15
But 4๐ 2 + 4๐ + 5 = 4 ๏ฟฝ๐ + 12๏ฟฝ2+ 4, hence
๐น (๐ ) =2๐
4 ๏ฟฝ๐ + 12๏ฟฝ2+ 4
+1
4 ๏ฟฝ๐ + 12๏ฟฝ2+ 4
=๐
2 ๏ฟฝ๐ + 12๏ฟฝ2+ 2
+14
1
๏ฟฝ๐ + 12๏ฟฝ2+ 1
=12
๐
๏ฟฝ๐ + 12๏ฟฝ2+ 1
+14
1
๏ฟฝ๐ + 12๏ฟฝ2+ 1
=12๐ + 1
2 โ12
๏ฟฝ๐ + 12๏ฟฝ2+ 1
+14
1
๏ฟฝ๐ + 12๏ฟฝ2+ 1
=12
๐ + 12
๏ฟฝ๐ + 12๏ฟฝ2+ 1
โ14
1
๏ฟฝ๐ + 12๏ฟฝ2+ 1
+14
1
๏ฟฝ๐ + 12๏ฟฝ2+ 1
=12
๐ + 12
๏ฟฝ๐ + 12๏ฟฝ2+ 1
(1)
Now we ready to do the inversion. Using ๐โ๐๐ก๐ (๐ก) โบ ๐น (๐ + ๐) and using sin (๐๐ก) โบ ๐๐ 2+๐2 , and
cos (๐๐ก)โบ ๐ ๐ 2+๐2 then
12
๐ + 12
๏ฟฝ๐ + 12๏ฟฝ2+ 1
โบ12๐โ
12 ๐ก cos (๐ก)
Hence
๐ (๐ก) =12๐โ
12 ๐ก cos (๐ก)
1.15 Section 6.3 problem 28
Find inverse Laplace transform of ๐น (๐ ) = 19๐ 2โ12๐ +3
Solution
19๐ 2 โ 12๐ + 3
=19
1๐ 2 โ 4
3 ๐ +13
=19
1
(๐ โ 1) ๏ฟฝ๐ โ 13๏ฟฝ
16
But1
(๐ โ 1) ๏ฟฝ๐ โ 13๏ฟฝ=
๐ด๐ โ 1
+๐ต
๐ โ 13
๐ด =
โโโโโโโโโโ
1
๏ฟฝ๐ โ 13๏ฟฝ
โโโโโโโโโโ ๐ =1
=32
๐ต = ๏ฟฝ1
(๐ โ 1)๏ฟฝ๐ = 1
3
= โ32
Hence
19๐ 2 โ 12๐ + 3
=19
โโโโโโโ32
1๐ โ 1
โ32
1๐ โ 1
3
โโโโโโโ (1)
Using
๐๐๐ก โบ1
๐ โ ๐Then (1) becomes
19๐ 2 โ 12๐ + 3
โบ19 ๏ฟฝ
32๐๐ก โ
32๐13 ๐ก๏ฟฝ
=16๐๐ก โ
16๐13 ๐ก
=16๏ฟฝ๐๐ก โ ๐
13 ๐ก๏ฟฝ
1.16 Section 6.3 problem 29
Find inverse Laplace transform of ๐น (๐ ) = ๐2๐โ4๐
2๐ โ1
solution
๐น (๐ ) =๐2
2๐โ4๐
๐ โ 12
Using
๐ข๐ (๐ก) ๐ (๐ก โ ๐)โบ ๐โ๐๐ ๐น (๐ ) (1)
Since1
๐ โ 12
โบ๐12 ๐ก
Then using (1)
๐โ4๐ 1
๐ โ 12
โบ๐ข4 (๐ก) ๐12 (๐กโ4)
17
Hence๐2
2๐โ4๐
๐ โ 12
โบ๐2
2๐ข4 (๐ก) ๐
12 (๐กโ4)
=12๐ข4 (๐ก) ๐
12 (๐กโ4)+2
=12๐ข4 (๐ก) ๐
12 ๐กโ2+2
=12๐ข4 (๐ก) ๐
๐ก2
Therefore
๐ (๐ก) =12๐ข4 (๐ก) ๐
๐ก2
ps. Book answer is wrong. It gives
๐ (๐ก) =12๐ข4 ๏ฟฝ
๐ก2๏ฟฝ ๐
๐ก2
1.17 Section 6.3 problem 30
Find Laplace transform of ๐ (๐ก) =
โงโชโชโจโชโชโฉ1 0 โค ๐ก < 10 ๐ก โฅ 1
solution
Writing ๐ (๐ก) in terms of Heaviside step function gives
๐ (๐ก) = ๐ข0 (๐ก) โ ๐ข1 (๐ก)
Using
๐ข๐ (๐ก) โบ ๐โ๐๐ 1๐
Therefore
โ{๐ข0 (๐ก)} = ๐โ0๐ 1๐ =1๐
โ {๐ข1 (๐ก)} = ๐โ๐ 1๐
Hence
โ{๐ข0 (๐ก) โ ๐ข1 (๐ก)} =1๐ โ ๐โ๐
1๐
=1๐ (1 โ ๐โ๐ ) ๐ > 0
1.18 Section 6.3 problem 31
Find Laplace transform of ๐ (๐ก) =
โงโชโชโชโชโชโชโจโชโชโชโชโชโชโฉ
1 0 โค ๐ก < 10 1 โค ๐ก < 21 2 โค ๐ก < 30 ๐ก โฅ 3
solution
18
Writing ๐ (๐ก) in terms of Heaviside step function gives
๐ (๐ก) = ๐ข0 (๐ก) โ ๐ข1 (๐ก) + ๐ข2 (๐ก) โ ๐ข3 (๐ก)
Using
๐ข๐ (๐ก) โบ ๐โ๐๐ 1๐
But ๐ (๐ก) = 1 in this case. Hence ๐น (๐ ) = 1๐ . Therefore
๐ (๐ก)โบ1๐ ๐0๐ โ
1๐ ๐โ๐ +
1๐ ๐โ2๐ โ
1๐ ๐โ3๐
=1๐ ๏ฟฝ1 โ ๐โ๐ + ๐โ2๐ โ ๐โ3๐ ๏ฟฝ ๐ > 0
1.19 Section 6.3 problem 32
Find Laplace transform of ๐ (๐ก) = 1 + โ2๐+1๐=1 (โ1)๐ ๐ข๐ (๐ก)
solution
Using
๐ข๐ (๐ก) โบ ๐โ๐๐ 1๐
Therefore
โ๏ฟฝ1 +2๐+1๏ฟฝ๐=1
(โ1)๐ ๐ข๐ (๐ก)๏ฟฝ = โ{1} +โ๏ฟฝ2๐+1๏ฟฝ๐=1
(โ1)๐ ๐ข๐ (๐ก)๏ฟฝ
=1๐ +
2๐+1๏ฟฝ๐=1
(โ1)๐1๐ ๐โ๐๐
=2๐+1๏ฟฝ๐=0
(โ1)๐1๐ ๐โ๐๐
=1๐
2๐+1๏ฟฝ๐=0
(โ๐โ๐ )๐
Since |๐โ๐ | < 1 the sum converges. Using โ๐0 ๐๐ = ๏ฟฝ
1โ๐๐+1
1โ๐๏ฟฝ. Where |๐| < 1. So the answer is
โ๏ฟฝ1 +2๐+1๏ฟฝ๐=1
(โ1)๐ ๐ข๐ (๐ก)๏ฟฝ =1๐
โโโโโ1 โ (โ๐โ๐ )2๐+2
1 โ (โ๐โ๐ )
โโโโโ
=1๐
โโโโโโ1 โ (โ๐)โ(2๐+2)๐
1 + ๐โ๐
โโโโโโ
Since 2๐ + 2 is even then
โ๏ฟฝ1 +2๐+1๏ฟฝ๐=1
(โ1)๐ ๐ข๐ (๐ก)๏ฟฝ =1๐ ๏ฟฝ1 + ๐โ(2๐+2)๐
1 + ๐โ๐ ๏ฟฝ ๐ > 0
1.20 Section 6.3 problem 33
Find Laplace transform of ๐ (๐ก) = 1 + โโ๐=1 (โ1)
๐ ๐ข๐ (๐ก)
solution
19
Using
๐ข๐ (๐ก) โบ ๐โ๐๐ 1๐
Therefore
โ๏ฟฝ1 +โ๏ฟฝ๐=1
(โ1)๐ ๐ข๐ (๐ก)๏ฟฝ = โ{1} +โ๏ฟฝโ๏ฟฝ๐=1
(โ1)๐ ๐ข๐ (๐ก)๏ฟฝ
=1๐ +
โ๏ฟฝ๐=1
(โ1)๐1๐ ๐โ๐๐
=1๐ +1๐
โ๏ฟฝ๐=1
(โ1)๐ ๐โ๐๐
=1๐ +1๐
โ๏ฟฝ๐=1
(โ๐โ๐ )๐
Butโ๏ฟฝ๐=1
๐๐ =๐
1 โ ๐|๐| < 1
Since ๐ > 0 then|๐โ๐ | < 1. So the answer is1๐ +1๐
โ๐โ๐
1 โ (โ๐โ๐ )=1๐ โ1๐
๐โ๐
1 + ๐โ๐
=1 + ๐โ๐ โ ๐โ๐
๐ (1 + ๐โ๐ )
=1
๐ (1 + ๐โ๐ )๐ > 0
1.21 Section 6.4 problem 21
August 7, 2012 21:04 c06 Sheet number 34 Page number 342 cyan black
342 Chapter 6. The Laplace Transform
Resonance and Beats. In Section 3.8 we observed that an undamped harmonic oscillator(such as a springโmass system) with a sinusoidal forcing term experiences resonance if thefrequency of the forcing term is the same as the natural frequency. If the forcing frequencyis slightly different from thenatural frequency,then the systemexhibits a beat. InProblems19 through 23 we explore the effect of some nonsinusoidal periodic forcing functions.
19. Consider the initial value problem
yโฒโฒ + y = f (t), y(0) = 0, yโฒ(0) = 0,where
f (t) = u0(t) + 2nโ
k=1(โ1)kukฯ(t).
(a) Draw the graph of f (t) on an interval such as 0 โค t โค 6ฯ.(b) Find the solution of the initial value problem.(c) Let n = 15 and plot the graph of the solution for 0 โค t โค 60. Describe the solutionand explain why it behaves as it does.(d) Investigate how the solution changes as n increases.What happens as n โ โ?
20. Consider the initial value problem
yโฒโฒ + 0.1yโฒ + y = f (t), y(0) = 0, yโฒ(0) = 0,where f (t) is the same as in Problem 19.(a) Plot the graph of the solution. Use a large enough value of n and a long enought-interval so that the transient part of the solution has become negligible and the steadystate is clearly shown.(b) Estimate the amplitude and frequency of the steady state part of the solution.(c) Compare the results of part (b) with those from Section 3.8 for a sinusoidally forcedoscillator.
21. Consider the initial value problem
yโฒโฒ + y = g(t), y(0) = 0, yโฒ(0) = 0,where
g(t) = u0(t) +nโ
k=1(โ1)kukฯ(t).
(a) Draw the graph of g(t) on an interval such as 0 โค t โค 6ฯ. Compare the graph withthat of f (t) in Problem 19(a).(b) Find the solution of the initial value problem.(c) Let n = 15 and plot the graph of the solution for 0 โค t โค 60. Describe the solutionand explain why it behaves as it does. Compare it with the solution of Problem 19.(d) Investigate how the solution changes as n increases.What happens as n โ โ?
22. Consider the initial value problem
yโฒโฒ + 0.1yโฒ + y = g(t), y(0) = 0, yโฒ(0) = 0,where g(t) is the same as in Problem 21.(a) Plot the graph of the solution. Use a large enough value of n and a long enough t-interval so that the transient part of the solution has become negligible and the steadystate is clearly shown.(b) Estimate the amplitude and frequency of the steady state part of the solution.
1.21.1 Part (a)
A plot of part (a) is the following
20
ฯ 2 ฯ 3 ฯ 4 ฯ 5 ฯ 6 ฯ
0.2
0.4
0.6
0.8
1.0
6.4 (21) part (a) plot
And a plot of part(a) for problem 19 is the following
ฯ 2 ฯ 3 ฯ 4 ฯ 5 ฯ 6 ฯ
-1.0
-0.5
0.5
1.0
6.4 (19) part (a) plot
We see the e๏ฟฝect of having a 2 inside the sum. It extends the step ๐ข๐ (๐ก) function to negative side.
1.21.2 Part (b)
The easy way to do this, is to solve for each input term separately, and then add all the solutions,since this is a linear ODE. Once we solve for the first 2-3 terms, we will see the pattern to use forthe overall solution. Since the input ๐ (๐ก) is ๐ข0 (๐ก) + โ
โ๐=1 (โ1)
๐ ๐ข๐๐ (๐ก), we will first first the response to๐ข0 (๐ก) , then for โ๐ข๐ (๐ก) then for +๐ข2๐ (๐ก), and so on, and add them.
When the input is ๐ข0 (๐ก), then its Laplace transform is 1๐ , Hence, taking Laplace transform of the
ODE gives (where now ๐ (๐ ) = โ๏ฟฝ๐ฆ (๐ก)๏ฟฝ)
๏ฟฝ๐ 2๐ (๐ ) โ ๐ ๐ฆ (0) + ๐ฆโฒ (0)๏ฟฝ + ๐ (๐ ) =1๐
Applying initial conditions
๐ 2๐ (๐ ) + ๐ (๐ ) =1๐
21
Solving for ๐0 (๐ ) (called it ๐0 (๐ ) since the input is ๐ข0 (๐ก))
๐0 (๐ ) =1
๐ ๏ฟฝ๐ 2 + 1๏ฟฝ
=1๐ โ
๐ ๐ 2 + 1
Hence
๐ฆ0 (๐ก) = 1 โ cos ๐ก
We now do the next input, which is โ๐ข๐ (๐ก), which has Laplace transform of โ ๐โ๐๐
๐ , therefore, followingwhat we did above, we obtain now
๐๐ (๐ ) =โ๐โ๐๐
๐ ๏ฟฝ๐ 2 + 1๏ฟฝ
= โ๐โ๐๐ ๏ฟฝ1๐ โ
๐ ๐ 2 + 1๏ฟฝ
The e๏ฟฝect of ๐โ๐๐ is to cause delay in time. Hence the the inverse Laplace transform of the above isthe same as ๐ฆ0 (๐ก) but with delay
๐ฆ๐ (๐ก) = โ๐ข๐ (๐ก) (1 โ cos (๐ก โ ๐))
Similarly, when the input is +๐ข2๐ (๐ก), which which has Laplace transform of ๐โ2๐๐
๐ , therefore, followingwhat we did above, we obtain now
๐๐ (๐ ) =๐โ2๐๐
๐ ๏ฟฝ๐ 2 + 1๏ฟฝ
= ๐โ2๐๐ ๏ฟฝ1๐ โ
๐ ๐ 2 + 1๏ฟฝ
The e๏ฟฝect of ๐โ2๐๐ is to cause delay in time. Hence the the inverse Laplace transform of the above isthe same as ๐ฆ0 (๐ก) but with now with delay of 2๐, therefore
๐ฆ2๐ (๐ก) = +๐ข2๐ (๐ก) (1 โ cos (๐ก โ 2๐))
And so on. We see that if we add all the responses, we obtain
๐ฆ (๐ก) = ๐ฆ0 (๐ก) + ๐ฆ๐ (๐ก) + ๐ฆ2๐ (๐ก) +โฏ= (1 โ cos ๐ก) โ ๐ข๐ (๐ก) (1 โ cos (๐ก โ ๐)) + ๐ข2๐ (๐ก) (1 โ cos (๐ก โ 2๐)) โโฏ
Or
๐ฆ (๐ก) = (1 โ cos ๐ก) +๐๏ฟฝ๐=1
(โ1)๐ ๐ข๐๐ (๐ก) (1 โ cos (๐ก โ ๐๐)) (1)
1.21.3 Part (c)
This is a plot of (1) for ๐ = 15
22
10 20 30 40 50 60t
-15
-10
-5
5
10
15
y(t)6.4 (21) part (c) plot
We see the solution growing rapidly, they settling down after about ๐ก = 50 to sinusoidal wave atamplitude of about ยฑ15. This shows the system reached steady state at around ๐ก = 50.
To compare it with problem 19 solution, I used the solution for 19 given in the book, and plottedboth solution on top of each others. Also for up to ๐ก = 60. Here is the result
problem 21
problem 19
2 ฯ 4 ฯ 6 ฯ 8 ฯ 10 ฯ 12 ฯ 14 ฯ 16 ฯ 18 ฯt
-30
-20
-10
10
20
30
y(t)
We see that problem 19 output follows the same pattern (since same frequency is used), but withdouble the amplitude. This is due to the 2 factor used in problem 19 compared to this problem.
1.21.4 Part(d)
At first, I tried it with ๐ = 50, 150, 250, 350, 450, 550. I can not see any noticeable change in the plot.Here is the result.
23
10 20 30 40 50 60
-20
-10
10
20
n = 50
10 20 30 40 50 60
-20
-10
10
20
n = 150
10 20 30 40 50 60
-20
-10
10
20
n = 250
10 20 30 40 50 60
-20
-10
10
20
n = 350
10 20 30 40 50 60
-20
-10
10
20
n = 450
10 20 30 40 50 60
-20
-10
10
20
n = 550
Even at ๐ = 2000 there was no change to be noticed.
10 20 30 40 50 60
-20
-10
10
20n = 2000
This shows additional input in the form of shifted unit steps, do not change the steady state solution.