{]- (1 = 42 X [-25-10) Ly 1 x(0) = 2 / y (0) = -6
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Hi, I found x(t) = 2cos(t)e-3t + 2sin(t)e-3t okay, but stuck on y(t). Could you please explain the steps?
!['X'
=
42
X
{][y
[-25-10) | y
X ₁ = AX
₁
42
-25-10
(4 (3-1)
-25
A = 42
1-25-10
(A-21) = 0
afcwming
10
ol
प
2
["₁²]- (-²+²) [60] [~] = [2
-25
-10
ol
V₂
Ò
V₁ =
7₁ (+)
[(-3+i) 0
0 (-3+1)
2
-0 (3-1)
2
1-12 +41
A
1
(12-4;)v, +2(√₂)=0
vi e
exit
= √2
Euler's formula:
-12 +41
x(0) = 2,
e
V₁
] [4=[8]
V₂
V₂ =
10
(4-2) (-10-2) -(2)/(25)=0
7² +62 +10=0
2₁ = −3+²₁ 22₂ = -3-i
= I
[J
V₁
[V]
V₂
=
0
0
(-3+i)t
✓
y(0) = -6
(-12² +4²) V₁_
2
, 3₁ = 7₁ = [24]
下
= cos (1 t)
ока рить
--12+4;
= cos + + ising.
_3+
+ i sin (it)
it](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d5b5049-d8dd-402a-aa3b-0cfd97dc82be%2F71b59542-6aae-4577-b3f5-99ab60357222%2Fgc5idnf_processed.png&w=3840&q=75)
Transcribed Image Text:'X'
=
42
X
{][y
[-25-10) | y
X ₁ = AX
₁
42
-25-10
(4 (3-1)
-25
A = 42
1-25-10
(A-21) = 0
afcwming
10
ol
प
2
["₁²]- (-²+²) [60] [~] = [2
-25
-10
ol
V₂
Ò
V₁ =
7₁ (+)
[(-3+i) 0
0 (-3+1)
2
-0 (3-1)
2
1-12 +41
A
1
(12-4;)v, +2(√₂)=0
vi e
exit
= √2
Euler's formula:
-12 +41
x(0) = 2,
e
V₁
] [4=[8]
V₂
V₂ =
10
(4-2) (-10-2) -(2)/(25)=0
7² +62 +10=0
2₁ = −3+²₁ 22₂ = -3-i
= I
[J
V₁
[V]
V₂
=
0
0
(-3+i)t
✓
y(0) = -6
(-12² +4²) V₁_
2
, 3₁ = 7₁ = [24]
下
= cos (1 t)
ока рить
--12+4;
= cos + + ising.
_3+
+ i sin (it)
it
Transcribed Image Text:2
7₁ (t) =
2 (os (t)
42 10s (+)
= (+) 'X
(₁.
x(0) =
=
2
-6
2
[-12 +41)
2 cos(t)
-12 cos (+)
11
e-34
e-ie
2 cos (t) + 2 i sin (t)
(12+41) (05 (+)
= -12 cost + 4i cost
e-3+
=
(los (It) + i sin (it) .e-
-34
+1
Lose -isin o
(os (it) - i sin (it)
(12+41) (isin (+))
- (12 i sin(t) + 4i-isin(t)))
= -12 isin (+) - 4i-i Sin(+)
+ C₂
C₁ cost)e-3t
C₁12 cos (f) ext
x (0) = C₁ 2 coste-²7 + C₂ 2 sint e-³t
2 sin (+)
-12 Sin (+) - 4 sin (f)
= -16 sin (4)
+
2 Sin(t)e-3+
е
- 16 Sin(t))
G₂ sin(t)e-3t
+(₂-16 Sin(t)e ²+
= 2
C₁ 2 cos (o) e° + (₂ 2 sinlo) e º = 2
C₁2(1)(1)
+ (₂2 (0) (0) =2
C₁ 2 =2
2
르
G-
- 12 cas(t)est +(₂-16 sin(t)e-³ = -6
(1) (-12) (cos(0)) e0 + (₁₂ (16)X(sin(6) eº = -6
(
e-36
C₁ = 1
_3t
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