Query_22

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course Mth 279

5/17

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:

The matrix is invertable when the determinate

det[A(t)] is not zero.

det[A(t)] = (t+1)(t+1) - t^2

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

= 2t + 1

Since this is not zero, then the matrix can be inverted

Since det[A(t)] cannot be zero.

2t + 1 = 0

t + 1/2 = 0

t = -1/2 The determinant would be zero here.

Therefore the matrix is invertable everywhere except

t = -1/2

[ t + 1, t; t, t + 1]^(-1) = 1/det[A(t)]*[ t + 1, -t; -t, t + 1]

=[ (t + 1)/(2t+1), -t/(2t+1); -t/(2t+1), (t + 1)/(2t+1)]

And as a check:

A(t)*A(t)^(-1) = I

[ t + 1, t; t, t + 1] * [ (t + 1)/(2t+1), -t/(2t+1); -t/(2t+1), (t + 1)/(2t+1)] = [1, 0; 0, 1]

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Good. Notice that the denominators are zero when t = -1/2 so that the inverse is not defined for t = -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:

If the limit of one in the matrix does not exist, then the limit of the matrix does not exist.

As t = 0

sin(t)/t = 0/0 = DNE

tcos(t) = 0

3/(t+1) = 3

e^(3t) = 1

sec(t) = 1

2t/(t^2-1) = 0

Since the limit of sin(t)/t does not exist,

the limit of the matrix does not exist.

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The limit of sin(t) / t as t -> 0 is 1, which is easily seen by l'Hopital's rule. The derivatives of both functions at t = 0 are 1, and both derivatives approach 1 as t approches 0.

Graphically the slope of y = sin(t) at t = 0 is 1, which is identical to the slope of y = t. So the graphs of sin(t) and t match at t = 0, and nearly match near t = 0. For any value of t near 0, therefore, one of these functions divided by the other will give you a result close to 1. And as t approaches zero, the limiting value is 1.

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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 ]

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Your solution:

The derivative of the matrix is the equal to the derivatives of all inside the matrix.

A'(t) = [ (sin(t))', (3 t)'; (t^2 + 2)' , (5)' ]

= [ cos(t), 3; 2t, 0 ]

A''(t) = [ (cos(t))', (3)'; (2t)', (0)' ]

=[ -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(t) = [ y1(t); y2(t)]

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

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

[y1'(t); y2'(t)] = [ t^2, 3; sin(t), t ][ y1(t); y2(t)] + [ 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:

The second derivative must be integrated twice to obtain A(t).

Which will result in some unknown constants C, and D.

A'(t) = [ t+C11, (1/2)t^2 + C12; C21, C22 ]

A(t) = [ (1/2)t^2 + tC11 + D11, (1/6)t^3 + tC12 + D12; tC21 + D21, tC22 + D22 ]

Using A(0) = [ 1, 1; -2, 1] we can solve for D due to all the other terms having t in them.

A(0) = [ D11, D12; D21, D22 ] = [ 1, 1; -2, 1]

A(t) = [ (1/2)t^2 + tC11 + 1, (1/6)t^3 + tC12 + 1; tC21 - 2, tC22 + 1 ]

Using A(1) = [-1, 2; -2, 3 ] we can solve for C.

A(1) = [ (1/2) + C11 + 1, (1/6) + tC12 + 1; C21 - 2, C22 + 1 ] = [-1, 2; -2, 3 ]

yeilding equations:

(1/2) + C11 + 1 = -1

(1/6) + C12 + 1 = 2

C21 -2 = -2

C22 + 1 = 3

C11 = -2.5

C12 = 5/6

C21 = 0

C22 = 2

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

to check this solution we can differentiate twice, and check to see if the second derivative matches the original we started with.

A'(t) = [ ((1/2)t^2 - 2.5t + 1)', ((1/6)t^3 + (5/6)t + 1)'; (-2)', (2t + 1)' ]

= [ t - 2.5, (1/2)t^2 + (5/6); 0, 2 ]

A''(t) = [ (t - 2.5)', ((1/2)t^2 + (5/6))'; (0)', (2)' ]

=[ 1, t; 0, 0 ]

This matches the original exactly proving that the solution is correct.

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

since, Int (e^s)ds = e^s

from 0 to t

= e^(t) - e^(0) = e^t - 1

Int (6s)ds = 3s^2

from 0 to t

= 3(t)^2 - 3(0)^2 = 3t^2

Int (cos(2 pi s))ds = sin(2 pi s)/(2 pi)

from 0 to t

= sin(2 pi (t))/(2pi) - sin(2 pi (0))/(2 pi) = sin(2 pi (t))/(2pi)

and Int(sin(2 pi s))ds = -cos(2 pi s)/(2 pi)

from o to t

= -cos(2 pi (t))/(2 pi) + cos(2 pi (0))/(2 pi)

= (-cos(2 pi t) + 1)/(2 pi)

= sin^2(pi t)/pi

A(t) = [ e^t - 1, 3t^2; sin(2 pi t)/(2pi), sin^2(pi s)/pi ]

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Given Solution:

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