Query_15

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course Mth 279

3/23 15

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:

Evaluate the wronksian at t = 3

W(3) = y_1(3) y_2Т(3) - y_2(3) y_1Т(3)

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

For a fundamental set the Wronskian is a non zero, so [y1, y2] is not a fundamental set.

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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 = 2 e^(-2 t) cos(t)

y1Т = 2 e^(-2t) -sin(t) - 4e^(-2t) cos(t)

y1Т = e^(-2t) (-2 sin(t) - 4 cos(t))

The second derivative will be done in pieces

d / dt e^(-2t) -sin(t)

= e^(-2t) - cos(t) - 2 e^(-2t) - sin(t)

= e^(-2t) (-cos(t) + 2 sin(t) This is eq. 1

d / dt - 2 e^(-2t) cos(t)

= -2 e^(-2t) - sin(t) - 4 e^(-2t) cos(t)

= e^(-2t) (2 sin(t) + 4 cos(t)) This is eq. 2

Adding eq. 1 and eq. 2 together we have

y1ТТ = 2 (3 cos(t) + 4 sin(t)) e^(-2t)

y1ТТ = e^(-2t) (6 cos(t) + 8 sin(t))

Now bringing y1, y1Т, and y1ТТ into the follow equation:

yТТ + 4 yТ + 5 y = 0

The e^(-2t) term can be divided out so I will ignore it.

0 = 6 cos(t) + 8 sin(t) + 4(-2 sin(t) - 4 cos(t)) + 5(2 cos (t)

0 = 6 cos(t) + 8 sin(t) - 8 sin(t) - 16 cos(t) + 10 cos(t)

0 = 0

y1 is a solution to the equation.

Now for the second equationЕ

y2 = e^(-2 t) sin(t)

y2Т = e^(-2 t) cos(t) - 2 e^(-2 t) sin(t)

y2Т = e^(-2 t) (cos(t) - 2 sin(t))

y2ТТ = d / dt e^(-2 t) cos(t) - 2 e^(-2 t) sin(t))

d/ dt e^(-2 t) cos(t)

= e^(-2 t) - sin(t) - 2 e^(-2 t) cos (t) eq 1

d / dt - 2 e^(-2 t) sin(t)

4 e^(-2 t) sin(t) - 2 e^(-2 t) cos(t) eq 2

Adding eq 1 and eq 2 back together we have

y2ТТ = e^(-2 t) (3 sin(t) - 4 cos(t))

Now bringing y1, y1Т, and y1ТТ into the follow equation:

yТТ + 4 yТ + 5 y = 0

0 = e^(-2 t) (3 sin(t) - 4 cos(t)) + 4 (e^(-2 t) (cos(t) - 2 sin(t)) + 5(e^(-2 t) sin(t))

0 = 3 sin(t) - 4 cos(t) + 4 cos(t) - 8 sin(t) + 5 sin(t)

0 = 0

y2 is a solution to the equation.

Initial conditions at t = 0 :

y1(0) = 2 e^0 cos (0) = 2

y1Т(0) = e^ 0 ( - 2 sin(0) - 4 cos(0)) = -4

y1ТТ(0) = e^0 (6 cos(0) + 8 sin(0)) - 6

y2(0) = e^ 0 ( sin(0)) = 0

y2Т(0) = e^ 0 (cos(0) - 2 sin(0)) = 1

y2ТТ(0) = e^ 0 ( 3 sin(0) - 4 cos(0)) = -4

If they are a fundamental set the wronksian will be a non-zero

y1 = f(t)

y2 = g(t)

W(t) = f(t) gТ(t) - g(t) fТ(t)

After a larrrrge amount of algebra this simplifies to

2 e^(-4t) which does not equal zero.

This indicates it is a fundamental set.

confidence rating #$&*:

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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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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:

To find out, I will compute the wronksian. If the wronksian is equal to zero, it is not a fundamental set.

W(t) = 2(y1 - y2)(y1-y2)Т - 2(y1-y2)Т(y1-y2)

W(t) = 0

Not 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):

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Self-critique rating:

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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:

The wronksian is

W(t) = [e^t , 2e^-t , sinh(t) ]

This is a long process detailed below:

(e^t)[(-2e^-t)(-sinh(t)0 - (2 e^-t)(cosh(t))] → Line 1

(-2 e^-t)[(e^t)(-sinh(t)) - (cosh(t)(e^t)] → Line 2

(sinh(t))[(e&t)(2e^-t) - (-2e^-t)(e^t)] → Line 3

W(t) = Line 1 - Line 2 + Line 3

Simplifying Line 1 we have:

[(-2)(-sinh(t)) - (2)(cosh(t)] = 2 ( sinh(t) - cosh(t) )

Simplifying Line 2 we have:

2 (-sinh(t) - cosh(t)) = -2(sinh(t) + cosh(t))

Simplifying Line 3 we have:

sinh(t)[2 + 2] = 4 sinh(t)

Finally, the wronksian is Line 1 - Line 2 + Line 3 which equals

4(sinh(t) - cosh(t))

This does not equal zero so they are a fundamental set.

@&
For a comparison, let's avoid the 2 on e^-t and do the Wronskian { [e^t, e^-t, sinh(t) }

Expressing the hyperbolic sines and cosines in terms of exponential functions our Wrokskian is

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 this is carefully expanded everything cancels out and we get 0.

Alternatively we could express the Wronskian as

det [ e^t, e^(-t), sinh(t);
e^t, -e^(-t), cosh(t);
e^t, e^(-t), sinh(t)]

= e^t * det( -e^-t, cosh(t); e^(-t), sinh(t) )
= e^t ( -e^(-t) sinh(t) - cosh(t) e^(-t) )
- e^(-t) (e^t sinh(t) - e^t( cosh(t) )
+ sinh(t) * (e^t * e^-t - (-e^-t) e^t) )

= -sinh(t) - cosh(t) - sinh(t) + cosh(t) + 2 sinh(t) * 0 = 0
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The only thing that changes if you use 2 e^-t instead of e^-t is that the determinant will be twice as great, for reasons that should be pretty clear.
*@

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The reason you get 0 is that sinh(t) = (e^t - e^(-t) ) / 2, which is a linear combination of e^t and 2 e^(-t) (being half the first minus a quarter of the second).
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Query_15

@& The only thing that changes if you use 2 e^-t instead of e^-t is that the determinant will be twice as great, for reasons that should be pretty clear. *@

@& The reason you get 0 is that sinh(t) = (e^t - e^(-t) ) / 2, which is a linear combination of e^t and 2 e^(-t) (being half the first minus a quarter of the second). *@

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

@&
The only thing that changes if you use 2 e^-t instead of e^-t is that the determinant will be twice as great, for reasons that should be pretty clear.
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@&
The reason you get 0 is that sinh(t) = (e^t - e^(-t) ) / 2, which is a linear combination of e^t and 2 e^(-t) (being half the first minus a quarter of the second).
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