Query 15

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course MTH 279

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:

We can say that they are not a fundamental set.

To be a fundamental set, the Wronskian must be nonzero at every t_0 in an interval.

W(t) = |{y1,y2};{y1’,y2’}|

W(3) = |{0,1};{0,2}| = 0 - 0 = 0

Since the Wronskian is ≠ 0, and there is no restriction on the domain of the solution, we know that y1 and y2 do not form a fundamental set.

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: 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’’ = 6e^(-2t)cos(t)+8e^(-2t)sin(t)

y2’ = -2e^(-2t)sin(t) + e^(-2t)cos(t)

y2’’ = 3e^(-2t)sin(t)-4e^(-2t)cos(t)

y1’’ + 4y1’ + 5y1 = 0 

y2’’ + 4y2’ + 5y2 = 0 

Plugging t=0 into y1,y1’,y2,y2’, we get the initial conditions:

y1(0) = 2, y1’(0) = -4;

y2(0) = 0, y2’(0) = 1;

W(t) = |{2e^(-2t)cos(t), e^(-2t)sin(t)};{ -4e^(-2t)cos(t)-2e^(-2t)sin(t), -2e^(-2t)sin(t) + e^(-2t)cos(t)} | = 2e^(-4t)

Since the Wronskian is non-zero for all values of t, then y1 and y2 do form a fundamental set.

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: 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:

Computing the Wronskian:

W(t) = |{2y1-2y2,y1-y2};{2y1’-2y2’,y1’-y2’}| = (2y_1y_1’-2y_1y_2’ - 2y_2y_1’+2y_2y_2’) - (2y_1’y_1-2y_1’y_2-2y_2’y_1+2y_2’y_2) = 0 for any and all t_0.

Also, I believe that if y_1_bar = 2y_2_bar, then they are linearly dependent, meaning that they cannot be a fundamental set.

confidence rating #$&*:

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Given Solution: Note that y_1_bar = 2 * y_2_bar.

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

y1,y2,and y3 are all continuous on the interval (-∞,∞). There are no restrictions on the domain of any of the solutions.

W(t) = |{y_1,y_2,y_3};{y_1’,y_2’,y_3’};{y_1”,y_2”,y_3”}| = y_1(y_2’y_3”-y_2”y_3’) - y_2(y_1’y_3”-y_1”y_3’) + y_3(y_1’y_2” - y_1”y_2’)

If we plug in our set and evaluate correctly, we should get W(t) = 0. This Wronskian is zero at any and all values of t, and since there are no restrictions on the domain of any of the solutions, we find that this may be a linear dependent set.

However, if we take pairwise Wronskians, we find that e^(t) and e^(-t) are linearly independent, we find that e^(t) and sinh(t) are linearly independent, and e^(-t) and sinh(t) are also linearly independent. This means that none of them are linearly dependent on each other, so they form a fundamental set.

@&

Being pairwise linearly independent doesn't make the entire set linearly independent.

y1 + y2 is clearly a linear combination of y1 and y2, y1 + y2 being a lienar combination of y1 and y2.

However if y1 and y2 are linearly independent, then y1 and y1+y2 must be linearly independent, and must y2 and y1 + y2.

*@

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"

&#Good work. See my notes and let me know if you have questions. &#