diffeq 3 systems of linear diffeq
DESCRIPTION
applied linear algebraTRANSCRIPT
7/21/2019 Diffeq 3 Systems of Linear Diffeq
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•
◦
◦
◦
•
◦
◦
•
◦ y + y = 0
∗ z = y w = y = z
y = −y
w = y = −y = −z
∗ y = z z = w w = −z
◦ y1 + y1 − y2 = 0 y2 + y1 + y2 sin(x) = ex
∗ z1 = y
1 z2 = y
2 z1 = y
1 = −y1 + y2 z2 = y
2 = ex − y1 −y2 sin(x) = ex − z1 − z2 sin(x)
∗ y1 = z1 y2 = z2 z1 = −y1 + y2 z2 =ex − z1 − z2 sin(x)
•
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• y1, · · · , yn
y1 = a1,1(x) · y1 + a1,2(x) · y2 + · · · + a1,n(x) · yn + p1(x)
y2 = a2,1(x) · y1 + a2,2(x) · y2 + · · · + a2,n(x) · yn + p2(x)
yn = an,1(x) · y1 + an,2(x) · y2 + · · · + an,n(x) · yn + pn(x)
ai,j(x) pj(x) 1 ≤ i, j ≤ n
◦ ai,j(x)
◦ p1(x), p2(x), · · · , pn(x)
◦ n y1(x0) = b1 y2(x0) = b2 . . .
yn(x0) = bn x0 x bi
•
n
• ai,j(x)
(y1
, y2
,· · ·
, yn)
y1 = a1,1(x) · y1 + a1,2(x) · y2 + · · · + a1,n(x) · yn
y2 = a2,1(x) · y1 + a2,2(x) · y2 + · · · + a2,n(x) · yn
yn = an,1(x) · y1 + an,2(x) · y2 + · · · + an,n(x) · yn
n
◦
◦
n
•
ai,j(x) pj(x) x = x0
y1 = a1,1(x) · y1 + a1,2(x) · y2 + · · · + a1,n(x) · yn + p1(x)
y2 = a2,1(x) · y1 + a2,2(x) · y2 + · · · + a2,n(x) · yn + p2(x)
yn = an,1(x) · y1 + an,2(x) · y2 + · · · + an,n(x) · yn + pn(x)
y1(x0) = b1 . . . yn(x0) = bn (y1, y2, · · · , yn)
x = x0
◦ y = ex · y +sin(x) · y z = 3x2 · y
y(a) = b1 z(a) = b2
• n s1 = (y1,1, y1,2, · · · , y1,n) s2 = (y2,1, y2,2, · · · , y2,n)
· · · sn = (yn,1, yn,2, · · · , yn,n)
W =
y1,1 y1,2 · · · y1,ny2,1 y2,2 · · · y2,n
yn,1 yn,2 · · · yn,n
◦ n s1, · · · , sn
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•
y1 = a1,1y1 + a1,2y2 + · · · + a1,nyn
y2 = a2,1y1 + a2,2y2 + · · · + a2,nyn
yn = an,1y1 + an,2y2 + · · · + an,nyn
• y = A·y y =
y1y2
yn
A =
a1,1 a1,2 · · · a1,na2,1 a2,2 · · · a2,n
an,1 an,2 · · · an,n
•
◦ v =
c1c2
cn
A λ y =
c1c2
cn
eλx
y = A · y
◦ y = eλx v x y = λeλxv = λ y = A · y
• A n v1, · · · , vn λ1, · · · , λn
y = A · y y = c1eλ1xv1 + c2eλ2xv2 + · · · + cneλnxvn
c1, · · · , cn
◦ A n v1, · · · , vn
λ1, · · · , λn A A = P −1DP
D λ1, · · · , λn P v1, · · · , vn
◦ eλ1xv1 eλ2xv2 · · · eλnxvn y = A · y
∗
W = e(λ1+···+λn)x ·
| | |
v1 · · · vn| | |
v1, · · · , vn eλ1xv1 eλ2xv2
· · · eλnxvn
∗ y = A · y
n
∗ n eλ1xv1 eλ2xv2
· · · eλnxvn n
∗ y = c1eλ1xv1 + c2eλ2xv2 + · · · +cneλnxvn c1, · · · , cn
•
A
◦ y = A · y n × 1 y n × n A
◦ A A A
n v1, · · · , vn
λ1, · · · , λn
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∗ λ
λ
λ
λ
◦ y = c1eλ1xv1 + c2eλ2xv2 + · · · + cneλnxvn
c1, · · · , cn
∗
∗ λ = a + bi v = w
1 + i w
2
λ = a − bi
v = w1 − i w2 v
∗ eλxv
eλxv eax( w1 cos(bx) − w2 sin(bx)) eax( w1 sin(bx) + w2 cos(bx))
◦ c1, · · · , cn
• y1 y2
y1 = y1 − 3y2y2 = y1 + 5y2
◦ y = A · y y =
y1y2
A =
1 −31 5
◦ A det(tI −A) = t − 1 3
−1 t − 5 = (t−1)(t−5)+3 = t2−6t+8
λ = 2, 4
∗ λ = 2
1 −31 5
·
a
b
=
2a
2b
a − 3b
a + 5b
=
2a
2b
a = −3b
−3
1
∗ λ = 2
1 −31 5
·
a
b
=
4a
4b
a − 3b
a + 5b
=
4a
4b
a = −b
−1
1
◦
y1y2 = c1
−31
· e2x + c2 −1
1 · e4x =
−3c1e2x + c2e4x
− c2e2x + c2e4x
• y1 y2
y1 = y2y2 = −y1
◦ y = A · y y =
y1y2
A =
0 1−1 0
◦ A det(tI − A) =
t −11 t
= t2 + 1
λ = ±i
∗ λ = i
0 1−1 0
·
a
b
=
ia
ib
b = ia
1
i
∗
λ = −i
λ = i 1
−i
◦
y1y2
= c1
1
i
· eix + c2
1−i
· e−ix
∗
1
i
· eix
1−i
· e−ix
∗ λ = i v =
10
+
01
i w1 =
10
w2 =
01
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∗
10
· cos(x) +
01
· sin(x) =
cos(x)
sin(x)
10
· sin(x) −
01
· cos(x) =
sin(x)− cos(x)
∗
y1y2
= c1
cos(x)
sin(x)
+c2
sin(x)− cos(x)
=
c1 cos(x) + c2 sin(x
c1 sin(x) − c2 cos(x
• y = A · y A n × n n
A
◦ v λ y = veλt
◦ y = A · y n
◦ A n
◦ A n
◦
y = A · y
◦ λ k λ k
λ k λ
• y = A · y
◦ n × n A
◦ λ n
v1, · · · , vk λ
∗ k = n
y = A · y eλtv1, · · · , eλtvn
∗ k < n w
v1, · · · , vk
(A − λI ) · w2 = w
(A − λI ) · w3 = w2
(A − λI ) · wl = wl−1.
eλt [ w] eλt[t w + w2] eλt[ t2
2 w + t w2 + w3]
· · · eλt tl−1l! w + tl−2
(l−2)! w2 + · · · + t wl−1 + wl
∗ λ
n n λ λ > 1
◦ y1, · · · , yn n
y = A · y y = c1y1 + c2 y2 + · · · + cn yn
∗
y y y y Re(y) Im(y) y
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•
y1 = 5y1 − 9y2y2 = 4y1 − 7y2
◦ y = A · y A =
5 −94 −7
∗ A−tI =
5 − t −9
4 −7 − t
det(A−tI ) = (5−t)(−7−t)−(4)(−9) = 1+2t+t2 = (t+1)2
∗ λ = −1
∗ λ = −1
6 −94 −6
·
a
b
=
00
2a − 3b = 0
a23a
32
◦ λ = −1
∗ w =
32
A − λI =
6 −94 −6
∗ w2 = a
b 6 −94 −6 ·
a
b = 32
·
2 −32 −3
11
2a − 3b = 1 a = 2 b = 1
· w2 =
21
∗
32
e−t
32
te−t +
21
e−t
◦
y1y2
= c1
32
e−t
+ c2
32
te−t +
21
e−t
∗ y
1 = (3c1 + 2c2 + 3c2t)e
−t
y2 = (2c1 + c2 + 2c2t)e−t
•
y1 = 4y1 − y2 − 2y3y2 = 2y1 + y2 − 2y3y3 = 5y1 − 3y3
◦ y = A · y A =
4 −1 −2
2 1 −25 0 −3
∗ A−tI =
4 − t −1 −2
2 1 − t −25 0 −3 − t
det(A−tI ) = 5
−1 −21 − t −2
+(3−t)
4 − t −12 1 − t
=
2 − t + 2t2 − t3 = (2 − t)(1 + t2)
∗ λ = 2, i, −i
∗ λ = 2 λ = 2
2 −1 −2
2 −1 −25 0 −5
·
a
b
c
=
0
00
2a−b−2c = 0 5a−5c = 0
c = a b = 2a − 2c = 0
a
0a
1
01
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∗ λ = i λ = i
4 − i −1 −2
2 1 − i −25 0 −3 − i
·
a
b
c
=
0
00
2 − i
4 − i −1 −2
−1 1 05 0 −3 − i
·
a
b
c
= 0
00
−a + b = 0
b = a
5a + (−3 − i)c = 0
c = 5
3 + ia =
5(3 − i)
10 a
a
a3 − i
2 a
2
23 − i
∗ λ = −i λ = −i λ = i
22
3 + i
.
◦
◦
y1
y2y3
= c1
1
01
e2t
+ c2
2
23 − i
eit
+ c3
2
23 + i
e−it
∗
y1
y2y3
= c1
1
01
e2t + c2
2 cos(t)
2 cos(t)3 cos(t) + sin(t)
+ c3
2sin(t)
2sin(t)− cos(t) + 3 sin(t)
∗
y1 = c1e2t + 2c2 cos(t) + 2c3 sin(t)y2 = 2c2 cos(t) + 2c3 sin(t)y3 = c1e2t + c2(3 cos(t) + sin(t)) + c3(− cos(t) + 3 sin(t))
•
y1 = 4y1 − y3y2 = 2y1 + 2y2 − y3y3 = 3y1 + y2
◦ y = A · y A =
4 0 −1
2 2 −13 1 0
∗ A −tI =
4 − t 0 −1
2 2 − t −13 1 −t
det(A −tI ) = (4−t)
2 − t −11 −t
+ (−1)
2 2 − t
3 1
=
8 − 12t + 6t2 − t3 = (2 − t)3
∗ λ = 2
∗ λ = 2
2 0 −1
2 0 −13 1 −2
·
a
b
c
=
0
00
2a − c = 0
3a + b − 2c = 0 c = 2a b = 2c − 3a = a
a
a
2a
∗
1
12
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◦ λ = 2
∗ w =
1
12
A − λI =
2 0 −1
2 0 −13 1 −2
∗ w2 =
a
b
c
2 0 −12 0 −1
3 1 −2 ·
a
b
c =
11
2
·
2 0 −1
2 0 −13 1 −2
112
2 0 −1
2 0 −13 1 −2
112
R2−R1−→
2 0 −1
0 0 03 1 −2
102
R3−2R1−→
2 0 −1
0 0 0−1 1 0
100
−a + b = 0 2a − c = 1 w2 w2 =
111
∗ w3 =
de
f
2 0 −12 0 −13 1 −2
·
de
f
=
111
·
2 0 −1
2 0 −13 1 −2
111
2 0 −1
2 0 −13 1 −2
111
R2−R1−→
2 0 −1
0 0 03 1 −2
101
R3−2R1−→
2 0 −1
0 0 0−1 1 0
10
−1
−a + b = −1
2a − c = 1
w3
w3 = 1
01
∗
1
12
e2t
1
12
te2t+
1
11
e2t
1
12
t2
2 e2t+
1
11
te2t+
1
01
e2t
◦ y1
y2y3
= c1
1
12
e2t
+ c2
1
12
te2t +
1
11
e2t
+ c3
1
12
t2
2 e2t +
1
11
te2t +
1
01
e2t
∗
y1 = ( c1 + c2 + c3 + c2t + c3t + 12c3t2)e2t
y2 = (
c1 +
c2 +
c2
t +
c3
t +
1
2c3
t2
)e2t
y3 = (2c1 + c2 + c3 + 2c2t + c3t + c3t2)e2t
•
◦
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◦
• 1× 1 y = ky y(0) = C
y(x) = ekxC
◦ n × n
◦ eA
• A n × n A eA
eA =
∞n=0
An
n!
◦ z
ez =
∞n=0
zn
n!
◦
◦ eA =
∞n=0
An
n! A
• A n × n y = A · y
y(0) = y0 y(x) = eAx · y0
◦
d
dx[eAx] eAx
A · eAx
◦ y(x) = eAx · y0
•
A
◦ P eP −1AP = P −1
eA
P
∗ eP −1AP =
∞n=0
(P −1AP )n
n! = P −1
∞n=0
An
n!
P = P −1
eA
P
(P −1AP )n = P (An)P
◦ D λ1, · · · , λn eD
eλ1 , · · · , eλn
∗
◦ A A = P −1DP D =
λ1
λn
eAx = P −1
eλ1x
eλnx
P
• eAx A =
0 −23 5
◦ A det(tI − A) = t(t − 5) + 6 = (t − 2)(t − 3) λ = 2, 3
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∗ λ = 2
0 −23 5
·
a
b
= 2
a
b
−2b
3a + 5b
=
2a
2b
a = −b
−b
b
λ = 2
−1
1
∗ λ = 3
0 −23 5
·
a
b
= 3
a
b
−2b
3a + 5b
=
3a
3b
a = −
2
3b
−2
3
b
b
λ = 3 −2
3
∗ A A = P −1DP
D =
2 00 3
P =
−1 −2
1 3
P −1 =
−3 −2
1 1
◦ eDx =
e2x 0
0 e3x
◦ eAx = P eDxP −1 =
−1 −2
1 3
e2x 0
0 e3x
−3 −2
1 1
=
3e2x − 2e3x 2e2x − 2e3
−3e2x + 3e3x −2e3x + 3e
• eJx
J =
λ 1
λ 1
1λ 1
λ
eAx A
A
◦ J = λI + N N
0 10 1
10 1
0
N n
◦ (Jx)k = xk(λI + N )k = λkI +k1
λk−1N 1 + · · · +
kk−n
λk−nN n +
· · · N n N n−1
eJx
◦ eJx =
eλx xeλx x2
2 eλx · · · xn−1
(n−1)!eλx
eλx xeλx
x2
2 eλx
eλx xeλx
eλx
, J n × n