Query 14

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course mth 279

7/16/13 9:21pm

Query 14 Differential Equations*********************************************

Question: Decide whether y_1 = 3 e^t, y_2 = e^(t + 3) are solutions to the equation y '' - y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (-1) = 1 and y ' (-1) = 0.

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Your solution:

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y_1 = 3 e^t

y_1' = 3 e^t

y_1"" = 3 e^t

y_2 = e^(t+3)

y_2' = e^(t+3)

y_2"" = e^(t+3)

in both cases when we plug values above into equation y"" - y = 0

the answer = 0, thus y_1 and y_2 are solutions.

Finding the wronskian results in W(t) = 0. Per Theorem 4.7 in the book this means that y_1 and y_2 are linearly dependent.

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Correct, and that is also because e^(t + 3) = e^3 * e^t, which is just a scalar multiple of the other function e^t.************

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Given Solution:

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Question: Decide whether y_1 = e^(-t) and y_2 = 2 e^(1 - t) are solutions to the equation y '' + 2 y ' + y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (0) = 1 and y ' (0) = 0.

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Your solution:

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y_1 = e^(-t)

y_1' = -e^(-t)

y_1"" = e^(-t)

y_2 = 2e^(1 - t)

y_2' = -2e^(1 - t)

y_2"" = 2e^(1 - t)

in both cases when we plug values above into equation y"" + 2 y' + y (not equal) 0

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Both of these functions are in fact solutions.

For example for y_1 = e^(-t) we have

y '' + 2 y ' + y = e^(-t) - 2 e^(-t) + e^(-t) = 0.

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thus y_1 and y_2 are not solutions.

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Given Solution:

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Question: Suppose y_1 and y_2 are solutions to the equation

y '' + alpha y ' + beta y = 0

and that y_1 = e^(2 t). Suppose also that the Wronskian is e^(-t).

What are the values of alpha and beta?

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Your solution:

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W(t) = e^(-t)

this means that the following is true:

y_2' * e^(2t) - y_2 * (2*e^(2t)) = e^(-t)

solving for y_1

y_2 = (1/2)*e^(-3t)

we can now write two equations using y_1 and y_2

-9 e^(-3t) + alpha ((3/2)*e^(-3t)) + beta ( (1/2) e^(-3t)) = 0

4 e^(2t) + alpha (2e^(2t) + beta (e^(2t)) = 0

solving system of equations we find

alpha = 22

beta = -48

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Given Solution:

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Good.

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&#This looks good. See my notes. Let me know if you have any questions. &#