Assignment 16

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course Mth 279

7/22

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:

1) sub in y_1 and y_2 into equation to see if they = 0

a)y’’_1 - y = (3e^t)’’ - 3e^t = 3e^t - 3e^t = 0

yes y_1 is solution

b) y’’_2 - y = (e^(t+3))’’ - e^(t+3) = e^(t+3) - e^(t+3) = 0

yes y_2 is solution

2) are solution linearly independent? use wronskian

W(t) = |f g; f’ g’|

W(t) = |3e^t e^(t+3); 3e^t e^(t+3)| = (3e^t)(e^(t+3) - (e^(t+3))(3e^t) = 0

since W(t) = 0 , they are NOT linearly independent

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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:

1) sub in y_1 and y_2

y_1 = e^-t ; y’_1 = -e^-t ; y’’_1 = e^-t

(e^-t) + 2(-e^-t) + e^-t = 0?

2e^-t - 2e^-t = 0 yes solution

y_2 = 2e^(1-t); y’_2 = -2e^(1-t) ; y’’_2 = 2e^(1-t)

(2e^(1-t)) + 2(-2e^(1-t)) + 2e^(1-t) = 0?

4e^(1-t) - 4e^(1-t) = 0 yes solution

2) linearly independent?

W(t) = | e^-t 2e^(1-t); -e^-t -2e^(1-t)| =

= (-2e^-t*e^(1-t)) - (-2e^-t*e^(1-t)) = 0

Not linearly independent

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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:

y_1 = e^(2t) ; y’_1 = 2e^(2t) ; y’’_1 = 4e^(2t)

1) Plug into equation

4e^(2t) + a(2e^(2t)) + b(e^(2t) = 0

2) factor out e^(2t)

e^(2t) (4 + 2a + b) = 0

divide e^(2t) over

4 + 2a + b = 0 (one soln - use wronskian to find y_2)

3) W(t) = |e^(2t) y_2; 2e^(2t) y’_2| = e^(2t)y’_2 - y_2(2e^(2t)) = e^-t

e^(2t)y’_2 - 2e^(2t)y_2 = e^-t

y’ - 2y = e^(-3t)

p(t) = -2 int(p(t)) = -2t

y_2 = Ce^(2t) + e^(2t)int(e^(-2t)*e^(-3t))

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You can integrate e^(-2t) e^(-3t), which is equal to e^(-5 t).

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This is as far as I could take the problem.

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