Query 22

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course MTH 279

5/3 2:08 pm

Query 22 Differential Equations*********************************************

Question:  Find the values for which the matrix

[ t + 1, t; t, t + 1]

pictured as:

is invertible.

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Your solution: 

invertible if det ≠ 0 --> (t+1)^2 - t^2 = 2t+1 --> det = 0 when t = -1/2

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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Question:  Find the limit as t -> 0 of the matrix

[ sin(t) / t, t cos(t), 3 / (t + 1); e^(3 t), sec(t), 2 t / (t^2 - 1) ]

pictured as

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Your solution: 

  [1, 0, 3; 1, 1, 0]

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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Question:  Find A ' (t) and A ''(t), where the derivatives are with respect to t and the matrix is

A = [ sin(t), 3 t; t^2 + 2, 5 ]

pictured as

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Your solution: 

 A'= [ cos t, 3; 2t, 0]

A''= [ -sin t, 0; 2, 0]

  

confidence rating #$&*:

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Given Solution: 

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Question:  Write the system of equations

y_1 ' = t^2 y_1 + 3 y_2 + sec(t)

y_2 ' = sin(t) y _1 + t y_2 - 5

in the form

y ' = P(t) y + g(t),

where P(t) is a 2 x 2 matrix and y and g(t) are 2 x 1 column vectors.

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Your solution: 

[y_1; y_2]' = [t^2, 3; sin t, t][ y_1; y_2] + [sec t; -5]

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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Question:  If

A '' = [1, t; 0, 0]

with

A(0) = [ 1, 1; -2, 1]

A(1) = [-1, 2; -2, 3 ]

then what is the matrix A(t)?

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Your solution: 

A' = [t+a1, 1/2t^2+ b1; c1, d1]

A = [ 1/2t^2 + a1t+a2, 1/6t^3 + b1t+b2; c1t + c2, d1t+d2]

a1 = -5/2, a2 = 1

b1= 5/6, b2 = 1

c1 = 0, c2 = -2

d1 = 2, d2 = 1

then

A= [ 1/2t^2 + -5/2t+1, 1/6t^3 + 5/6t+1; -2, 2t+1]

 

 Confidence rating:

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Given Solution: 

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Question:  Find the matrix A(t), defined by

A(t) = integral( B(s) ds, s from 0 to t),

where

B = [ e^s, 6s; cos(2 pi s), sin(2 pi s) ].

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Your solution: 

 integrate B

[ e^s, 3s^2; 1/2π sin 2πs, -1/2π cos 2πs]

evaluate

A = [e^t -1, 3t^2; (1/2π sin(2π*t)), (-1/2π cos (2π*t)+ 1/2π)]

 Confidence rating:

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Given Solution: 

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Self-critique (if necessary):

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&#Very good responses. Let me know if you have questions. &#