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course Mth 279
April 30 around 3:15pm.
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:
y1 = 3 e^t
y2 = e^(t + 3)
y '' - y = 0 (main equation)
Plug in y1 for all y’s in main equation:
(3e^t)’’ - (3e^t) = 0
= 0, therefore y1 IS a solution.
Plug in y2 for all y’s in main equation:
(e^(t+3))’’ - (e^(t + 3)) = 0
= 0, therefore y2 is ALSO a solution.
To test for linearly independence, use the equation:
cf(x) + kg(x) = 0 (c and k are constants)
c (3e^t) + k(e^(t+3)) = 0
3ce^t + ke^(t+3) = 0
This is where I’m stuck….do I plug in y (-1) = 1 and y ' (-1) = 0 at this point?!
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& To check for linear independence you can calculate the Wronskian.
You will find that the Wronskian is zero, so the solutions are not linearly independent.*@
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Question: Decide whether y_1 = e^(-t) and y_2 = 2 e^(1 - t) are solutions to the equation y '' + 2 y ' + y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (0) = 1 and y ' (0) = 0.
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Your solution:
y1 = e^(-t)
y2 = 2 e^(1 - t)
y '' + 2 y ' + y = 0 (main equation)
Plug in y1 for all y’s in main equation:
(e^(-t))’’ - (e^(-t)) = 0
= 0, therefore y1 IS a solution.
Plug in y2 for all y’s in main equation:
(2 e^(1 - t))’’ - (2 e^(1 - t))= 0
= 0, therefore y2 is ALSO a solution.
To test for linearly independence, use the equation:
cf(x) + kg(x) = 0 (c and k are constants)
c (e^(-t)) + k(2 e^(1 - t))= 0
ce^(-t) + 2ke^(1-t) = 0
Just like the previous problem, this is where I’m stuck….do I plug in y (0) = 1 and y ' (0) = 0 at this point?!
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& Find the Wronskian. I believe you'll find these solutions not to be a fundamental set; the Wronskian will 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?
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Your solution:
y '' + alpha y ' + beta y = 0
W(t) = e^(-t)
W(t) = f(t)g’(t) - g(t)f’(t)
e^(-t) = f(t)g’(t) - g(t)f’(t)
e^(-t) = [g’(t)*e^(2t)] - [g(t)*2e^(2t)]
Need to solve for g(t) and g’(t)
That will then tell you what alpha and beta is in the “second order linear homogeneous” equation: y '' + alpha y ' + beta y = 0
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& The Wronskian is the determinant of the matrix
[ y_1, y_2; y_1 ' , y_2 ' ]
which in this case could be expresses as
det [ f, g; f ' , g ' ] = f g ' - f ' g.*@
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Self-critique (if necessary):
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Self-critique rating:
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Self-critique (if necessary):
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