Query_26

#$&*

course MTH-279

08/03, around 11:50PM

Query 26 Differential Equations*********************************************

Question: Find the solutions to y ' = A y when

A = [ 5, 3; -4, -3 ]

and

y(1) = [2; 0].

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

p(lambda) = det[A-lambda I]

lambda = [5, 3; -4, -3] = [lambda, 0; 0, lambda]

A - lambda I = [5, 3; -4, -3] - [lambda, 0; 0, lambda] = [5-y, 3-0; -4-0, -3-lambda] = [5-y, 3; -4, -3-lambda]

det[A - lambda I] = det[5-y, 3; -4, -3-lambda]

= [(5-lambda) * (-3-lambda)] - [(-4 * 3)]

=[-15 - 5lambda + 3lambda + lambda^2] + 12

-15 - 2lambda + lambda^2 + 12 -> -3 -2lambda + lambda^2

lambda^2 - 2 lambda - 3

(lambda + 1)(lambda - 3)

lambda = -1, 3

= det[5-lambda, 3; -4, -3-lambda] -> [5+1, 3; -4, -3+1] = [6, 3; -4, -2]

use [A - lambda I] * [x1; x2] = [0;0]

[6, 3 ; -4, -2] * [x1; x2] = [0,0]

6x1 + 3x2 = 0

-4x1 - 2x2 = 0

-----

[6,3; -4, 2]

divide by 6 -> [1, ½; -4, -2]

R2 + 4R1 -> [1, ½; 0,0] -> [1, ½; 0,0] * [x1; x2] = [0;0]

x1 + ½ x2 = 0

x1 = -1/2x2

-2x1 = x2

x = [1, -2]

---

det [5-lambda, 3; -4, -3-lambda] -> [5-3, 3; -4, -3-3] = [2, 3; -4, -6]

use A - lambda I * [x1; x2] = [0;0]

[2, 3; -4, -6] * [x1; x2] = [0;0]

[2, 3; -4, -6]

divide by 2 -> [1, 3/2; -4, -6]

R2 + 4R1 -> [1, 3/2; 0;0] -> [1, 3/2; 0;0] * [x1; x2] = [0;0]

x1 + 3/2x2 = 0

x1 = -3/2x2

x2 = 2, x1 = -3

----

(lambda1, lambda2) = 3(3, x2 vector) , x2vector = [-3, 2]

y2(t) = e^3t * [-3;2] = [-3e^3t; 2e^3t]

----

y(t) = C1y1(t) + C2y2(t)

= [e^-t, -3e^3t; -2e^-t, 2e^3t] * [C1; C2]

plug in initial value(1)

W(t) = ((e^-t * 2e^3t) - ((-2e^-t * -3et))

= (2e^2t) - (6e^3t) -> (2e^2(1)) - (6e^2(1))

= 2e^2 - 6e^2

= -4e^2 -> det.

W(t) does not equal zero

---

plug in initial value (1)

= [e^-1, -3e^3; -2e^-1, 2e^3] * [C1; C2] = [2;0]

[C1; C2] = [e^-1, -3e^3; -2e^-1, 2e^3]^-1 * [2;0]

= 1-4e^2 [2e^3, 3e^3; 2e^-1, e^-1] * [0;2]

=[-e/2, -3e/4; -1/2e^3, -1/4e^-3] * [2;0]

=[-e+0; -e^-3 + 0] = [-e;-e^-3] = [C1; C2]

---

y(t) = C1y1(t) + C2y2(t)

y(t) = e^-1 [e^-t; -2e^-t] - e^-3 [ -3e^3t; 2e^3t]

 [e^t-1; 2e^-t+1] + {3e^3t-3; -2e^3t-3]

= [-e^t-1 + 3e^3t-3; 2e^-t+1 - 2e^3t-3] -> y(t)

Confidence rating: 2,3

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

Self-critique rating: OK

Query_26

#$&*

course MTH-279

08/03, around 11:50PM

Query 26 Differential Equations*********************************************

Question: Find the solutions to y ' = A y when

A = [ 5, 3; -4, -3 ]

and

y(1) = [2; 0].

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

p(lambda) = det[A-lambda I]

lambda = [5, 3; -4, -3] = [lambda, 0; 0, lambda]

A - lambda I = [5, 3; -4, -3] - [lambda, 0; 0, lambda] = [5-y, 3-0; -4-0, -3-lambda] = [5-y, 3; -4, -3-lambda]

det[A - lambda I] = det[5-y, 3; -4, -3-lambda]

= [(5-lambda) * (-3-lambda)] - [(-4 * 3)]

=[-15 - 5lambda + 3lambda + lambda^2] + 12

-15 - 2lambda + lambda^2 + 12 -> -3 -2lambda + lambda^2

lambda^2 - 2 lambda - 3

(lambda + 1)(lambda - 3)

lambda = -1, 3

= det[5-lambda, 3; -4, -3-lambda] -> [5+1, 3; -4, -3+1] = [6, 3; -4, -2]

use [A - lambda I] * [x1; x2] = [0;0]

[6, 3 ; -4, -2] * [x1; x2] = [0,0]

6x1 + 3x2 = 0

-4x1 - 2x2 = 0

-----

[6,3; -4, 2]

divide by 6 -> [1, ½; -4, -2]

R2 + 4R1 -> [1, ½; 0,0] -> [1, ½; 0,0] * [x1; x2] = [0;0]

x1 + ½ x2 = 0

x1 = -1/2x2

-2x1 = x2

x = [1, -2]

---

det [5-lambda, 3; -4, -3-lambda] -> [5-3, 3; -4, -3-3] = [2, 3; -4, -6]

use A - lambda I * [x1; x2] = [0;0]

[2, 3; -4, -6] * [x1; x2] = [0;0]

[2, 3; -4, -6]

divide by 2 -> [1, 3/2; -4, -6]

R2 + 4R1 -> [1, 3/2; 0;0] -> [1, 3/2; 0;0] * [x1; x2] = [0;0]

x1 + 3/2x2 = 0

x1 = -3/2x2

x2 = 2, x1 = -3

----

(lambda1, lambda2) = 3(3, x2 vector) , x2vector = [-3, 2]

y2(t) = e^3t * [-3;2] = [-3e^3t; 2e^3t]

----

y(t) = C1y1(t) + C2y2(t)

= [e^-t, -3e^3t; -2e^-t, 2e^3t] * [C1; C2]

plug in initial value(1)

W(t) = ((e^-t * 2e^3t) - ((-2e^-t * -3et))

= (2e^2t) - (6e^3t) -> (2e^2(1)) - (6e^2(1))

= 2e^2 - 6e^2

= -4e^2 -> det.

W(t) does not equal zero

---

plug in initial value (1)

= [e^-1, -3e^3; -2e^-1, 2e^3] * [C1; C2] = [2;0]

[C1; C2] = [e^-1, -3e^3; -2e^-1, 2e^3]^-1 * [2;0]

= 1-4e^2 [2e^3, 3e^3; 2e^-1, e^-1] * [0;2]

=[-e/2, -3e/4; -1/2e^3, -1/4e^-3] * [2;0]

=[-e+0; -e^-3 + 0] = [-e;-e^-3] = [C1; C2]

---

y(t) = C1y1(t) + C2y2(t)

y(t) = e^-1 [e^-t; -2e^-t] - e^-3 [ -3e^3t; 2e^3t]

 [e^t-1; 2e^-t+1] + {3e^3t-3; -2e^3t-3]

= [-e^t-1 + 3e^3t-3; 2e^-t+1 - 2e^3t-3] -> y(t)

Confidence rating: 2,3

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

Self-critique rating: OK

#*&!

Query_26

#$&*

course MTH-279

08/03, around 11:50PM

Query 26 Differential Equations*********************************************

Question: Find the solutions to y ' = A y when

A = [ 5, 3; -4, -3 ]

and

y(1) = [2; 0].

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

p(lambda) = det[A-lambda I]

lambda = [5, 3; -4, -3] = [lambda, 0; 0, lambda]

A - lambda I = [5, 3; -4, -3] - [lambda, 0; 0, lambda] = [5-y, 3-0; -4-0, -3-lambda] = [5-y, 3; -4, -3-lambda]

det[A - lambda I] = det[5-y, 3; -4, -3-lambda]

= [(5-lambda) * (-3-lambda)] - [(-4 * 3)]

=[-15 - 5lambda + 3lambda + lambda^2] + 12

-15 - 2lambda + lambda^2 + 12 -> -3 -2lambda + lambda^2

lambda^2 - 2 lambda - 3

(lambda + 1)(lambda - 3)

lambda = -1, 3

= det[5-lambda, 3; -4, -3-lambda] -> [5+1, 3; -4, -3+1] = [6, 3; -4, -2]

use [A - lambda I] * [x1; x2] = [0;0]

[6, 3 ; -4, -2] * [x1; x2] = [0,0]

6x1 + 3x2 = 0

-4x1 - 2x2 = 0

-----

[6,3; -4, 2]

divide by 6 -> [1, ½; -4, -2]

R2 + 4R1 -> [1, ½; 0,0] -> [1, ½; 0,0] * [x1; x2] = [0;0]

x1 + ½ x2 = 0

x1 = -1/2x2

-2x1 = x2

x = [1, -2]

---

det [5-lambda, 3; -4, -3-lambda] -> [5-3, 3; -4, -3-3] = [2, 3; -4, -6]

use A - lambda I * [x1; x2] = [0;0]

[2, 3; -4, -6] * [x1; x2] = [0;0]

[2, 3; -4, -6]

divide by 2 -> [1, 3/2; -4, -6]

R2 + 4R1 -> [1, 3/2; 0;0] -> [1, 3/2; 0;0] * [x1; x2] = [0;0]

x1 + 3/2x2 = 0

x1 = -3/2x2

x2 = 2, x1 = -3

----

(lambda1, lambda2) = 3(3, x2 vector) , x2vector = [-3, 2]

y2(t) = e^3t * [-3;2] = [-3e^3t; 2e^3t]

----

y(t) = C1y1(t) + C2y2(t)

= [e^-t, -3e^3t; -2e^-t, 2e^3t] * [C1; C2]

plug in initial value(1)

W(t) = ((e^-t * 2e^3t) - ((-2e^-t * -3et))

= (2e^2t) - (6e^3t) -> (2e^2(1)) - (6e^2(1))

= 2e^2 - 6e^2

= -4e^2 -> det.

W(t) does not equal zero

---

plug in initial value (1)

= [e^-1, -3e^3; -2e^-1, 2e^3] * [C1; C2] = [2;0]

[C1; C2] = [e^-1, -3e^3; -2e^-1, 2e^3]^-1 * [2;0]

= 1-4e^2 [2e^3, 3e^3; 2e^-1, e^-1] * [0;2]

=[-e/2, -3e/4; -1/2e^3, -1/4e^-3] * [2;0]

=[-e+0; -e^-3 + 0] = [-e;-e^-3] = [C1; C2]

---

y(t) = C1y1(t) + C2y2(t)

y(t) = e^-1 [e^-t; -2e^-t] - e^-3 [ -3e^3t; 2e^3t]

 [e^t-1; 2e^-t+1] + {3e^3t-3; -2e^3t-3]

= [-e^t-1 + 3e^3t-3; 2e^-t+1 - 2e^3t-3] -> y(t)

Confidence rating: 2,3

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

Self-critique rating: OK

#*&!#*&!

&#Your work looks good. Let me know if you have any questions. &#