2) Given that p"(v) + 2p'(v) - 4p(v) = 8v

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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11. 1.
8+8²
=> P(s) = ²(s²+2s-4)
2
By decomposition we can write
Cs+D
s²-2s-4
2)
Given that
p"(v) + 2p'(v) - 4p(v) = 8v
Taking Laplace transform we get
(s²P(s) - sp(0) - p'(0)) + 2(sP(s) - p(0)) - 4P(s) = (s² +2s - 4)P(s) − 1 =
=> (s² + 2s - 4)P(s) = 8+²
P(s) = A ++
=> 8+ s² = As³ - 2As2-4As + Bs2 - 2Bs - 4B
+Cs³ +Ds²
=> 8+ s² = s³(A+C) + s²(B - 2A + D)
+(-4A - 2B)s - 4B
So, we get
P(s) = -2+2-3154
Step 3/3
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⇒A+C =0, B - 2A + D = 1,-4A - 2B = 0, -4B = 8
=> B = -2, A = 1, C = -1,D = 5
As we get
P(s) = ---- (51)²-5 + (8-1)²-5
Taking inverse Laplace we get
p(v) = 1-2v- e'cos ((√5)v) +
This is the Required
I.
4e sin((√5)v)
√5
4
5
A
Transcribed Image Text:O Search (Alt+Q) ings Review Help ou need to edit, it's safer to stay in Protected View. to this PC V View 11. 1. 8+8² => P(s) = ²(s²+2s-4) 2 By decomposition we can write Cs+D s²-2s-4 2) Given that p"(v) + 2p'(v) - 4p(v) = 8v Taking Laplace transform we get (s²P(s) - sp(0) - p'(0)) + 2(sP(s) - p(0)) - 4P(s) = (s² +2s - 4)P(s) − 1 = => (s² + 2s - 4)P(s) = 8+² P(s) = A ++ => 8+ s² = As³ - 2As2-4As + Bs2 - 2Bs - 4B +Cs³ +Ds² => 8+ s² = s³(A+C) + s²(B - 2A + D) +(-4A - 2B)s - 4B So, we get P(s) = -2+2-3154 Step 3/3 Enable Editing ⇒A+C =0, B - 2A + D = 1,-4A - 2B = 0, -4B = 8 => B = -2, A = 1, C = -1,D = 5 As we get P(s) = ---- (51)²-5 + (8-1)²-5 Taking inverse Laplace we get p(v) = 1-2v- e'cos ((√5)v) + This is the Required I. 4e sin((√5)v) √5 4 5 A
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T
I
T
1
1
=> G(s) —
*
S
(³) G(s) = 8² 149
s2
+
30+ = 1
49
=> C= 147
58
1
Given function is
g(t)-3g(u)du = cos (7t)
Taking Laplace transform we get
L-¹(g(t)) — 3L ¹(ſ'g(u)du) = L−¹(cos (7t))
3G(s)
S
B = ⇒ B= 49
58
So, we get
G(s) = 9
I
=> G(s) = (s−3)(s² | 49)
Using partial decomposition we can write
Bs + C
G(s) = 3
58(s-3)
A = 1-B⇒ A= 8
9
S
s²49
+
.
9
=> G(s) = 58(8-3)
S²49
=> s² = A (s² +49) + (Bs + C)(s − 3)
=> s² = s² (A + B) + s(C − 3B) + (49A — 3C)
A+B=1
C-3B = 0 ⇒ B=
49A3C = 0⇒ A= 3C
49
K
+
2
49s-+-147
s²1-49
49s
s²149
1
+
Taking inverse Laplace we get
9e3t
g(t)
58
I
147
s² +49
1
+49cos (7t) + 21sin (7t)
3
E
Enable Ed
|
F
Transcribed Image Text:Mailings Review View Help Unless you need to edit, it's safer to stay in Protected View. T I T 1 1 => G(s) — * S (³) G(s) = 8² 149 s2 + 30+ = 1 49 => C= 147 58 1 Given function is g(t)-3g(u)du = cos (7t) Taking Laplace transform we get L-¹(g(t)) — 3L ¹(ſ'g(u)du) = L−¹(cos (7t)) 3G(s) S B = ⇒ B= 49 58 So, we get G(s) = 9 I => G(s) = (s−3)(s² | 49) Using partial decomposition we can write Bs + C G(s) = 3 58(s-3) A = 1-B⇒ A= 8 9 S s²49 + . 9 => G(s) = 58(8-3) S²49 => s² = A (s² +49) + (Bs + C)(s − 3) => s² = s² (A + B) + s(C − 3B) + (49A — 3C) A+B=1 C-3B = 0 ⇒ B= 49A3C = 0⇒ A= 3C 49 K + 2 49s-+-147 s²1-49 49s s²149 1 + Taking inverse Laplace we get 9e3t g(t) 58 I 147 s² +49 1 +49cos (7t) + 21sin (7t) 3 E Enable Ed | F
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