query 15

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course Mth 279

7/21 10:44 pm

Query 15 Differential Equations*********************************************

Question:  Suppose y1 and y2 are solutions to y '' + 2 t y ' + t^2 y = 0.  If y1(3) = 0, y1 ' (3) = 0, y2(3) = 1 and y2 ' (3) = 2, can you say whether {y1, y2} is a fundamental set?  If so, is it or isn't it?

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Your solution: 

I can evaluate the Wronskian at t = 3

 W(3) = [ 0 1,

0 2]

W(3) = 0(2) - 1(0) = 0

{y1, y2} is a fundamental set because at t=3, the Wronskian is zero.

 

@&

Good work on the Wronskian.

However for a fundamental set the Wronskian is nonzero.

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Given Solution: 

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Question:  Are y1 = 2 e^(-2 t) cos(t) and y2 = e^(-2 t) sin(t) solutions to the equation

y '' + 4 y ' + 5 y = 0?

What are the initial conditions at t = 0?

Is {y1, y2} a fundamental set?

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Your solution: 

 y1' = -4e^(-2t)cos(t) + 2e^(-2t)sin(t)

 y1'' = -8e^(-2t) cos(t) + 4e^(-2t) - 4e^(-2t)sin(t) +2e^(-2t) cos(t)

Plugging in:

[-8e^(-2t) cos(t) + 4e^(-2t) - 4e^(-2t)sin(t) +2e^(-2t) cos(t) ] + 4[-4e^(-2t)cos(t) + 2e^(-2t)sin(t)] + 5[e^(-2t)cos(t)] = 0

e^(-2t)[-8cos(t) + 2cos(t) -16cos(t) + 10 cos(t) +4sin(t) - 4sin(t) + 8sin(t) ] = 0

e^(-2t)(-12cos(t) + 8sin(t) ) does not equal 0

y1 is not a solution to the equation.

y2' = -2e^(-2t)sin(t) + e^(-2t) cos(t)

y2'' = 4e^(-2t) sin(t) - 2e^(-2t)cos(t) - 2e^(-2t)cos(t) - e^(-2t)sin(t)

plugging in:

[4e^(-2t) sin(t) - 2e^(-2t)cos(t) - 2e^(-2t)cos(t) - e^(-2t)sin(t)] + 4[-2e^(-2t)sin(t) + e^(-2t) cos(t)] + 5[e^(-2t)sin(t)] = 0

e^(-2t)[4sin(t) - sin(t) - 8 sin(t) + 5sin(t) - 2cos(t) - 2cos(t) + 4 cos(t) ] = 0

e^(-2t) [0] = 0

0=0

y2 is a solution.

I don't think this is right, but I could not find my error in testing y1

 

 

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Given Solution: 

 

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Question:  y1_bar = 2 y1 - 2 y2 and y2_bar = y1 - y2.  Is {y1_bar, y2_bar} a fundamental set?

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Your solution: 

 

 if y1 = 2y2

W = [y1 2y1,

y1' 2y1']

2*y1*y1' - 2*y1*y1' = 0

Yes, it is a fundamental set

 

 

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Given Solution:  Note that y_1_bar = 2 * y_2_bar. 

 

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Question:  Is {e^t, 2 e^(-t), sinh (t) } a fundamental set on the interval (-infinity, infinity)?

 

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Your solution: 

 yes because none of the solutions have asymptotes.

 What is the difference between sinh and sin???

 

 

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sinh(t) = (e^t - e^(-t)) / 2 and

cosh(t) = (e^t + e^(-t)) / 2

The Wronskian of the 3-element set {y_1, y_2, y_3} is det ( [ y_1, y_2, y_3; y_1 ' , y_2 ', y_3 '; y_1 '', y_2 '', y_3 ''] ).

Using sinh(t) = (e^t - e^(-t) ) / 2 (and perhaps noting that this function is a linear combination of the first two, which does not bode well for a nonzero Wronskian) we have

det [ e^t, e^(-t), (e^t - e^(-t)) / 2; e^t, -e^(-t), (e^t + e^(-t)) / 2; e^t, e^(-t), (e^t - e^(-t)) / 2 ].

If we carefully expand this determinant you will find that all the terms pair up and cancel, giving the expected result that the determinant is zero.

On t > 0, | t | = t so the two functions y = t and y = | t | are identical, hence not linearly independent.

On -infinity < t < infinity, however, the equation

c_1 t + c_2 | t | = 0

would imply that for t > 0

c_1 t + c_2 t = 0,

and hence that c_2 = - c_1;

however for t < 0 our equation would imply that

c_1 t + c_2 ( -t ) = 0,

hence that c_2 = c_1.

Unless c_1 = c_2 = 0, this is a contradiction and we conclude that the two functions are linearly independent.

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confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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