Query 14

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course MTH 279

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:

Taking the first and second derivatives of each solution and substituting them into the differential equation shows that they are both solutions to the equation.

{} denotes a row, ; denotes a row break

W(t) = |{3e^t,e^(t+3)};{3e^t,e^(t+3)}| = 0

C(3e^t)+K(e^(t+3))=0

3Ce^(t)+Ke^(t)*e^(3)=0

(3C+Ke^3)e^t=0

3C+Ke^3 = 0

There are non-zero combinations of C and K that will yield 0 (such as C = 1, K = 3e^(-3)), therefore, the solutions are linearly dependent.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

Taking the first and second derivatives of y_1 and y_2 and substituting them back into the differential equation shows that they are both solutions to the equation.

W(t) = |{e^(-t),2e^(1-t)};{-e^(-t),-2e^(1-t)}| = 0

C(e^(-t))+K(2e^(1-t)) = 0

Ce^(-t) + 2Ke^(1)*e^(-t) = 0

(C+2Ke^(1)) e^(-t) = 0

C+2Ke^1 =0

There are non-zero combinations of C and K that will yield 0 (such as C=1 and K=e^(-1)/2), therefore, the solutions are linearly dependent.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

W(t) = |{e^(2t), y_2};{2e^(2t),y_2’}| = e^(2t)y_2’ - 2e^(2t)y_2 = e^(-t)

e^(2t)y_2’ - 2e^(2t)y_2 = e^(-t) >> y’ - 2y = e^(-3t)

µ = e^(ʃ-2dt) = e^(-2t)

e^(-2t)y’ - 2e^(-2t)y = e^(-5t)

ʃd/dt(e^(-2t)y)dt = ʃe^(-5t)dt

e^(-2t)y = -e^(-5t)/5 + C

y = -e^(-3t)/5+Ce^(2t)

y’ = 3e^(-3t)/5+Ce^(2t)

W(t) = e^(2t)*(3e^(-3t)/5+Ce^(2t))-2e^(2t)(3e^(-3t)/5+Ce^(2t))=e^(-t)

C = 0

y_2 = -e^(-3t)/5

y_2’ = 3e^(-3t)/5

y_2’’ = -9e^(-3t)/5

y_2’’+αy_2’+βy_2 = 0

β = 3α-9

y_1 = e^(2t)

y_1’ = 2e^(2t)

y_1’’ = 4e^(2t)

y_1’’+αy_1’+ (3α-9)y_1 = 0

α = 1

β = 3-9 = -6

y’’ + y’ - 6y = 0

y_1=e^(2t), y_2= -e^(-3t)/5

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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