Query 22

#$&*

course MTH 279

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:

Invertibility depends on the determinant of the matrix being non-zero.

Taking the determinant, we have:

(t+1)(t+1)-t^2 = 2t+1

Thus, the determinant is non-zero for all values of t ≠ -1/2.

The matrix is invertible for {t|t≠-1/2}.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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): OK

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Self-critique rating: OK

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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’(t) = [cos(t), 3; 2t, 0]

A”(t) = [-sin(t), 0; 2, 0]

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK

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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’ = [y_1’;y_2’]

p(t) = [t^2, 3; sin(t), t]

y = [y_1;y_2]

g(t) = [sec(t); -5]

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK

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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) = ʃA”dt = [ʃdt, ʃtdt; ʃ0dt, ʃ0dt] = [t+C, t^2/2+C; C, C]

Imposing A’(1) = [-1, 2; -2, 3],

A’(t) = [t-2, t^2/2 + 3/2; -2, 3]

A(t) = ʃA’dt = [t^2/2-2t+C, t^3/6 + 3t/2 + C; -2t+C, 3t+C]

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

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

A(t) = ʃB(s)ds from 0 to t = [e^s+C|(0,t), 3s^2+C|(0,t); sin(2πs)/(2π)+C|(0,t), -cos(2πs)/(2π)+C|(0,t)]

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK"

&#Good responses. Let me know if you have questions. &#