Query_15

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course MTH-279

07/29 around 10:20PM.Just turning in some homework done leading up to test two, of which I have taken already.

Query 15 Differential Equations

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Question: Suppose y1 and y2 are solutions to y '' + 2 t y ' + t^2 y = 0. If y1(3) = 0, y1 ' (3) = 0, y2(3) = 1 and y2 ' (3) = 2, can you say whether {y1, y2} is a fundamental set? If so, is it or isn't it?

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Your solution:

W(t) = f(t)g’(t) - f’(t)g(t)

Using the sets given -> (0 * 2) - (0 * 1) = 0

Not a set and linearly dependent since W = 0

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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Question: Are y1 = 2 e^(-2 t) cos(t) and y2 = e^(-2 t) sin(t) solutions to the equation

y '' + 4 y ' + 5 y = 0?

What are the initial conditions at t = 0?

Is {y1, y2} a fundamental set?

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Your solution:

use product rule in this problem

y1 = 2e^-2t cost(t)

y1’ = 2(-2e^-2t cost(t)) - (e^-2t sin(t))

-4e^-2t cos(t) - 2e^-2t sin(t)

e^-2t (-4cos(t) - 2sin(t))

-----

y1’’ = -2e^-2t (-4cos(t) - 2sin(t)) + e^-2t (4sin(t) - 2cos(t))

8e^-2t cos(t) + 4e^-2t sin(t) + 4e^-2t sin(t) - 2e^-2t cos(t)

6e^-2t cos(t) + 8e^-2t sin(t)

-16e^-2t cos(t) - 8e^-2t sin(t)

+10 e^-2t cos(t)

----

6cos(t) + 8sin(t)

-16cos(t) - 8sin(t)

+10 cos(t)

-----

0 = 0

-----

y2 = e^-2t sin(t)

y2’ = -2e^-2t (sint) + e^-2t (cost)

y2’’ = 4e^-2t(sint) - 2e^-2t(cost) +2e^-2t(cost) - e^-2t(sint)

= 3e^-2t(sint) - 4e^-2t(cost)

------

3e^-2t(sint) - 4e^-2t(cost)

-8e^-2t(sint) + 4e^-2t(cost)

+5e^-2t(sint)

-----

0=0

----

t = 0?

y1 = 2e^-2t

y1 = 2(1) * 1

y1 (0) =2

-----

y1’ = e^-2t(-4cost - 2sint)

y1’ = 1(-4 - 0)

y1’ (0) = -4

----

y1’’ = 6e^-2t cost + 8e^-2t sint

y1’’ = 6 + 0

y1’’ (0) = 6

-----

y2 = e^-2t sint

y2 = 1 * 0

y2 (0) = 0

----

y2’ = -2r^-2t sint + e^-2t cost

Y2’ = 0 + 1

y2’ (0) = 1

----

y2’’ = 3e^-2t sint -4e^-2t cost

y2’’ =0 - 4

y2’’ (0) = -4

-----

Using the W(t) = method

(2 * 1) - (4 * 0)

2 + 0

2 = 0

is a fundamental set because W does not equal zero

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: y1_bar = 2 y1 - 2 y2 and y2_bar = y1 - y2. Is {y1_bar, y2_bar} a fundamental set?

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Your solution:

Using the W(t) = method

(2y1 - 2y2)(y1-y2)’ - (y1-y2)(2y1-2y2)’

2(y1-y2)(y1-y2)’ - 2(y1-y2)(y1-y2)’ = 0

0 = 0

not a fundamental set

confidence rating #$&*:

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

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Given Solution: Note that y_1_bar = 2 * y_2_bar.

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

Self-critique (if necessary): OK

Self-critique rating: OK

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Question: Is {e^t, 2 e^(-t), sinh (t) } a fundamental set on the interval (-infinity, infinity)?

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Your solution:

Confidence rating:

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Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

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Question: y1_bar = 2 y1 - 2 y2 and y2_bar = y1 - y2. Is {y1_bar, y2_bar} a fundamental set?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Using the W(t) = method

(2y1 - 2y2)(y1-y2)’ - (y1-y2)(2y1-2y2)’

2(y1-y2)(y1-y2)’ - 2(y1-y2)(y1-y2)’ = 0

0 = 0

not a fundamental set

confidence rating #$&*:

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

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

Given Solution: Note that y_1_bar = 2 * y_2_bar.

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

Self-critique (if necessary): OK

Self-critique rating: OK

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Question: Is {e^t, 2 e^(-t), sinh (t) } a fundamental set on the interval (-infinity, infinity)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Confidence rating:

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Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

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Question: y1_bar = 2 y1 - 2 y2 and y2_bar = y1 - y2. Is {y1_bar, y2_bar} a fundamental set?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Using the W(t) = method

(2y1 - 2y2)(y1-y2)’ - (y1-y2)(2y1-2y2)’

2(y1-y2)(y1-y2)’ - 2(y1-y2)(y1-y2)’ = 0

0 = 0

not a fundamental set

confidence rating #$&*:

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

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

Given Solution: Note that y_1_bar = 2 * y_2_bar.

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

Self-critique (if necessary): OK

Self-critique rating: OK

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Question: Is {e^t, 2 e^(-t), sinh (t) } a fundamental set on the interval (-infinity, infinity)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Confidence rating:

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

"

Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

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