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course Mth 279
5/4
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:
Pluging in each solution to the equation varifies that they are solutions:
y_1(t) = 3e^t
y_1'(t) = 3e^t
y_1''(t) = 3e^t
y '' - y = 0
3e^t - 3e^t = 0
0=0
y_2(t) = e^(t + 3)
y_2'(t) = e^(t + 3)
y_2''(t) = e^(t + 3)
y '' - y = 0
e^(t + 3) - e^(t + 3) = 0
0=0
Then using the Wronskian, we can determine if its a fundamental set, and whether its Linearly Independant.
W(t) = |f(t) g(t)| =f(t)g'(t) - g(t)f '(t)
|f '(t) g'(t)|
W(t) = |3e^t e^(t+3)| = (3e^t)e^(t+3) - (3e^t)e^(t+3) = 0
|3e^t e^(t+3)|
Since the Wronskian equals zero, they are not a fundamental set, and therefore linearly dependant.
confidence rating #$&*:
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Given Solution:
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Question: Decide whether y_1 = e^(-t) and y_2 = 2 e^(1 - t) are solutions to the equation y '' + 2 y ' + y = 0.
If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution,
as well as a particular solution for which y (0) = 1 and y ' (0) = 0.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Pluging in each solution to the equation varifies that they are solutions:
y_1(t) = e^(-t)
y_1'(t) = -e^(-t)
y_1''(t) = e^(-t)
y '' + 2 y ' + y = 0
e^(-t) - 2 e^(-t) + e^(-t) = 0
-e^(-t) + e^(-t) = 0
0=0
y_2(t) = 2 e^(1 - t)
y_2'(t) = -2 e^(1 - t)
y_2(t) = 2 e^(1 - t)
y '' + 2 y ' + y = 0
2e^(1 - t) - 4e^(1 - t) + 2e^(1 - t) = 0
-2e^(1 - t) + 2e^(1 - t) = 0
0=0
Then using the Wronskian, we can determine if its a fundamental set, and whether its Linearly Independant.
W(t) = |f(t) g(t)| =f(t)g'(t) - g(t)f '(t)
|f '(t) g'(t)|
W(t) = |e^(-t) 2e^(1 - t)| = e^(-t)*(-2e^(1 - t)) - (2e^(1 - t))*(-e^(-t)) = 0
|-e^(-t) -2e^(1 - t)|
Since the Wronskian equals zero, they are not a fundamental set, and therefore linearly dependant.
confidence rating #$&*:
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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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Good.
Side note:
e^(1 - t) = e^1 * e^(-t) = e * e^(-t), which is a constant multiple of e^(-t). So this function is clearly linearly dependent on e^(-t).
You can see how if one function is a multiple of the other, the determinant will have to be zero.
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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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Pluging in each solution to the equation varifies that they are solutions:
y_1(t) = e^(2 t)
y_1'(t) = 2e^(2 t)
y_1''(t) = 4e^(2 t)
y'' + Ay' +By = 0
4e^(2 t) + A(2e^(2 t)) + Be^(2 t) = 0
e^(2 t) * (2A + B + 4) = 0
2A + B + 4 = 0
This is one solution of 2 needed to find A, and B.
Using W(t), we can find y_2(t)
W(t) = |f(t) g(t)| =f(t)g'(t) - g(t)f '(t)
|f '(t) g'(t)|
W(t) = |e^(2t) y_2(t)| = e^(-t)
|2e^(2t) y_2'(t)|
e^(2t) * y_2'(t) - y_2(t) * 2e^(2t) = e^(-t)
(y_2(t) * e^(2t))' = e^(-t)
integration of both sides yeilds:
y_2(t) * e^(2t) = -e^(-t)
y_2(t) = -e^(-3t)
y_2'(t) = 3e^(-3t)
y_2''(t) = -9e^(-3t)
y'' + Ay' +By = 0
-9e^(-3t) + A(3e^(-3t)) +B(-e^(-3t)) = 0
e^(-3t) * (3A - B - 9) = 0
3A - B - 9 = 0
This becomes our second equation needed to find A, and B.
2A + B + 4 = 0
3A - B - 9 = 0
adding these 2 equations yeilds:
5A - 5 = 0
A = 1
Plugging A back in to get B.
2A + B + 4 = 0
2(1) + B + 4 = 0
B = -6
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:"
Self-critique (if necessary):
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Very good job.
See my one note for a useful observation.
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