Query 14

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course MTH 279

6:10 pm 3/31 Seems query 13 was a repeat of some of query 12's questions

Query 14 Differential Equations*********************************************

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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:

Simple substitution confirms these are both solutions but the Wronskian is:

3e^t(e^(t+3))-3e^t(e^(t+3)) = 0

this means they aren't linearly independent and in fact

x(3e^t) = e^(t+3) where x = 1/3*e^3

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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:

Substitution confirms these are solutions

y= e^-t, y'=-e^-t, y''= e^-t

y= 2e^(1-t), y'= -2e^(1-t), y''= 2e^(1-t)

W= e^-t*(-2e^(1-t))-(-e^-t)*2e^(1-t) = 0

So they're not lin ind and

2e^1 *(e^(-t))= 2e^(1-t)

confidence rating #$&*:8232;Given Solution:

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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:

y1= e^2t, y1'= 2e^2t, y1''=4e^2t

4e^2t+alpha*2e^2t+beta*e^2t =0

4+2alpha+beta= 0

W= e^-t = e^2t * y2' - y2*2e^2t

y'2-2y2= e^-3t

So y2 must be multiple of e^-3t

Ae^-3t - 2Ae^-3t = e^-3t

A = -1

y2 = -e^-3t, y2'= 3e^-3t, y2''= -9e^-3t

-9+3alpha-beta =0

4+2alpha+beta= 0

alpha = 1, beta =-6

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Given Solution:

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GOod. Check my notes.

04-01-2011

&#Your work looks very good. Let me know if you have any questions. &#

course MTH 279

2:28 pm 4/1

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

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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:

Use Wronskian

det [[0,1][0,2]]= 0(2)-1(0)= 0

so not fundamental set

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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)

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

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

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

Factor out e^(-2 t) and combine like terms

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

The two parenthesis both = 0 so y1 is a solution. The book just has e^(-2 t) cos(t) but the 2 is consistent throughout and any constant multiple would work. We can easily see that a similar process for y2 would yield the same results all we have to do is replace sin and cos and arrange some signs.

y2= e^-2t sin t

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

sin t: 4-1-8+5 = 0

cos t: -2 -2 +4 = 0

Conditions at t=0:

y1(0)= 2, y1'(0)=-4, y1''(0)= 6

y2(0)= 0, y2'(0)=1, y2''(0)= -4

det [[y1, y2][y1', y2']] = 2e^-2t

This only equals 0 at t = inf

since t cannot equal, but only approach infinity, it follows that the solution is never zero. Good thing, because if it was zero at one point it would be zero everywhere.

8iConfidence rating: 2.9

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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:

Form matrix det:

det [[2, -2][1, -1]] = -2+2 =0

So not a fundamental set

Seems kind of obvious since 2*(y2) = y1

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

Show that the functions y1 = t and y2 = | t | are linearly dependent on the interval t > 0, but are linearly independent on the interval -infinity < t < infinity.

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Your solution:

All three expressions are defined on the interval.

Can't do a 2x3 det but

det [[e^t, 2e^-t][e^t, -2e^-t]] =-4

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

det [[sinh t, e^t][cosh t, e^t]] = 1

so they are linearly independent

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.

y1=t & y2 = |t| for t > 0 are clearly equivalent since |t| = t for t > 0

However, for -inf< t< inf the expressions are not = since y2(-t)= |-t| = t, y(-t)= -t

On t > 0, | t | = t so the two functions 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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Given Solution:

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