Query_15

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course Mth 279

5/4

Query 15 Differential Equations*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Using the Wronskian we can determine this:

W(t) = |f(t) g(t)| = f(t)g'(t) - g(t)f '(t)

|f '(t) g'(t)|

Substitute f(t) with y1(3) and g(t) with y2(3) as well as with their corresponding derivatives.

W(t) = |0 1| = 0(2) - 1(0) = 0

|0 2|

Since the Wronskian, W(t) = 0 these are not a fundamental set of Equations.

confidence rating #$&*:

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2

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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: --------------------------------------------------------------

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

y1 ' (t) = (d/dt)(2e^(-2t)cos(t))

= 2 ((e^(-2t))(-sin(t)) + (-2e^(-2t))(cos(t)))

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

y1 '' (t) = 2 * (d/dt)((e^(-2t))(-sin(t)) + (-2e^(-2t))(cos(t)))

Easier to do this one in pieces:

(d/dt)((e^(-2t))(-sin(t))

=(e^(-2t))(-cos(t)) + (-2e^(-2t))(-sin(t))

=(e^(-2t))(-cos(t) + 2sin(t))

(d/dt)(-2e^(-2t))(cos(t))

=(-2e^(-2t))(-sin(t)) + (-2)(-2e^(-2t))(cos(t))

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

Putting the whole equation back together yeilds:

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

y1 '' (t) = (e^(-2t))(8sin(t) + 6cos(t))

Now that we have all 3 this solution can be checked with the original equation:

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

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

y1 '' (t) = (e^(-2t))(8sin(t) + 6cos(t))

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

(e^(-2t))(8sin(t) + 6cos(t)) + 4 (e^(-2t))(-2sin(t) - 4cos(t)) + 10e^(-2t)cos(t) = 0

(e^(-2t)) * (8sin(t) + 6cos(t) - 4(2sin(t) + 4cos(t)) + 10cos(t) = 0

8sin(t) + 6cos(t) - 8sin(t) - 16cos(t) + 10cos(t) = 0

8sin(t) - 8sin(t) = 0

0 = 0

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Now we do the same with the second solution:

y2 = e^(-2 t) sin(t)

y2 ' (t) = (d/dt)(e^(-2t)sin(t))

= (e^(-2t))(cos(t)) + (-2e^(-2t))(sin(t))

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

y2 '' (t) = (d/dt)((e^(-2t))(cos(t)) + (-2e^(-2t))(sin(t)))

Easier to do this one in pieces:

(d/dt)((e^(-2t))(cos(t))

=(e^(-2t))(-sin(t)) + (-2e^(-2t))(cos(t))

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

(d/dt)(-2e^(-2t))(sin(t))

=(-2e^(-2t))(cos(t)) + (-2)(-2e^(-2t))(sin(t))

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

Putting the whole equation back together yeilds:

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

y2 '' (t) = (e^(-2t))(3sin(t) - 4cos(t))

Now that we have all 3 this solution can be checked with the original equation:

y2(t) = e^(-2t)sin(t)

y2 ' (t) = (e^(-2t))(cos(t) - 2sin(t))

y2 '' (t) = (e^(-2t))(3sin(t) - 4cos(t))

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

(e^(-2t))(3sin(t) - 4cos(t)) + 4(e^(-2t))(cos(t) - 2sin(t)) + 5e^(-2t)sin(t) = 0

(e^(-2t))(3sin(t) - 4cos(t) + 4(cos(t) - 2sin(t)) + 5sin(t)) = 0

(3sin(t) - 4cos(t) + 4cos(t) - 8sin(t) + 5sin(t)) = 0

(3sin(t) - 8sin(t)) + 5sin(t)) = 0

8sin(t) - 8sin(t) = 0

Therefore both are solutions!

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Initial conditions:

y1(0) = 2e^(-2*0)cos0 = 2

y1 ' (0) = (e^(-2*0))(-2sin(0) - 4cos(0)) = -4

y1 '' (0) = (e^(-2*0))(8sin(0) + 6cos(0)) = 6

y2(0) = e^(-2*0)sin(0) = 0

y2 ' (0) = (e^(-2*0))(cos(0) - 2sin(0)) = 1

y2 '' (0) = (e^(-2*0))(3sin(0) - 4cos(0)) = -4

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To find out if they are a fundamental set we must use the wronskian.

W(t) = |f(t) g(t)| =f(t)g'(t) - g(t)f '(t)

|f '(t) g'(t)|

W(t) = |2e^(-2t)cos(t) e^(-2t)sin(t) |

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

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

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

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

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

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

=(2e^(-4t))

Since its not zero, they should be a fundamental set.

confidence rating #$&*:

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3

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Given Solution:

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Self-critique (if necessary):

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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:

W(t) = |f(t) g(t)| =f(t)g'(t) - g(t)f '(t)

|f '(t) g'(t)|

W(t) = |2 y1 - 2 y2 y1 - y2 | = 2(y1 - y2)(y1 - y2)' - 2(y1 - y2)(y1 - y2)' = 0

|(2 y1 - 2 y2)' (y1 - y2)' |

Since the wronskian, W(t) = 0 they are not a fundamental set.

@&

Good.

Since y_1_bar is just double y_2_bar we would expect that it would turn out that way.

*@

confidence rating #$&*:

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3

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Given Solution: Note that y_1_bar = 2 * y_2_bar.

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Self-critique (if necessary):

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Self-critique rating:

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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:

W(t) = |f(t) g(t)| =f(t)g'(t) - g(t)f '(t)

|f '(t) g'(t)|

W(t) = |e^t 2e^(-t) sinh(t)| =f(t)g'(t) - g(t)f '(t)

|e^t -2e(-t) cosh(t)|

= |2e^(-t) sinh(t)| -|e^t sinh(t)| +|e^t 2e^(-t) |

|-2e(-t) cosh(t)| |e^t cosh(t)| |e^t -2e^(-t) |

= (2e^(-t))cosh(t) + (2e(-t))sinh(t) - (e^t)cosh(t) + (e^t)sinh(t) + (e^t)(-2e^(-t)) - 2e^(-t)e^t

= (2e^(-t))cosh(t) + (2e(-t))sinh(t) - (e^t)cosh(t) + (e^t)sinh(t) - 4

= (2e^(-t))(e^t + e^(-t))/2 + (2e(-t))(e^t - e^(-t))/2 - (e^t)(e^t + e^(-t))/2 + (e^t)(e^t - e^(-t))/2 - 4

= 1 + e^(-2t) + 1 - e^(-2t) - (1/2)(e^(2t) + 1) + (1/2)(e^(2t) - 1) - 4

= 2 - (1/2)(e^(2t) + 1 + e^(2t) - 1) - 4

= 2 - 1 - 4 = -3

Because its not zero, they are a fundamental set.

confidence rating #$&*:

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

3

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

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