query 14

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course Mth 279

4/28 5

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:

Sub y(t) = e^rt where y'' = r^2e^rt, now put back into equation

r^2e^rt - e^rt = 0

factor out e, where e cannot = 0

so you are left with

r^2 - 1 = 0

r = +-1

so the equation y(t) = c_1e^t + c_2e^-t

solutions to the equation are not 3e^t or the other.

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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:

again using the same process, sub y = e^rt and factor out the e^rt to be left with

r^2 +2r + 1 = 0

this factors to (r+1)(r+1) = 0 where r = -1 for both . so our solution to the equation is now y(t) = c_1e^-t which is in fact y_1 = e^-t. now to determine if they are linearly independent.....

here it only meets the y_1 solution, but when I put in the y_2 and y_1 into a matrix [ y_1 y_2 , y_1' y_2' ] the determinant is actually 0, and this goes against the wronskian which has to equal a non-zero determinant,

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but if the determinent is 0, then that means it is linearly dependent. so no, it is not linearly independent

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Right. Good.

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confidence rating #$&*:

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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:

if the W is e^-t,

then e^-t = e^2t + ( y_2) where y_2 would = -e^3t

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

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