Assignment 24

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course Mth 279

8/5

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:

calculate determinate

= ((t+1)(t+1))-(t*t)

= t^2+2t+1-t^2

=2t+1

find values of t that make det[A] =/0

0=/2t+1

t=/(-1/2)

Therefore A is invertible as long as t doesNOT equal -1/2

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Given Solution:

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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:

take limit of each entry

=|lim sin(t)/t , 0 , 3; 1, 1, 0|

do L’H of sin(t)/t = cos(t)/1 =1

so lim A(t) = |1 ,0,3; 1,1,0|

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Given Solution:

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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 = |sin(t), 3t; t^2+2, 5|

A’(t) = |cos(t), 3; 2t, 0|

A”(t) = |-sin(t), 0; 2,0|

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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’ = [y’1; y’2]

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

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

[y’1; y’2] = [t^2, 3; sin(t),t]*[y1;y2] + [sec(t); -5]

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Given Solution:

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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:

1) take integral twice of A”

A(t) = [t^2/2 + C_11t+D_11, t^3/6+C_12t+D_12; C_21t+D_21, C_22t+D_22]

A(0) = [1,1;-2,1] = [D_11, D_12; D_21, D_22]

A(1) = [-1,2;-2,3] = [1/2+C_11(1)+1, 1/6+C_12(1)+1; C_21+(-2), C_22(1)+1]

-1 = 1/2+ C_11+1; C_11 = -2.5

2= 1/6+C_12+1 ; C_12 = 5/6

-2 = C_21-2; C_21=0

3= C_22+1; C_22=2

A(t) = [ t^2/2 -2.5t+1, t^3/6+5/6t_1; -2, 2t+1]

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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:

A(t) = int(B(s))from 0 to t

A(t) = [e^t - e^0, 3t^2-3(0)^2; 1/(2pi)*sin(2pit)-1/(2pi)*sin(2pi(0)), -1/(2pi)*(cos(2pi(t)-cos(2pi(0)))

= [e^t-1, 3t^2; 1/(2pi)*sin(2pit), -1/(2pi)*(cos(2pi*t)-1)]

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. &#