Query 27

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course MTH 279

4/19 7 pm

Query 27 Differential Equations*********************************************

Question: Find the eigenvalues of the matrix [3, 1; -2, 1] and find the corresponding eigenvectors.

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Your solution:

(2+ i , [ -(1+i)/2 ; 1] ) and (2- i , [ -(1-i)/2 ; 1] )

evals I found using det[ (3 - lam)(1 - lam) +2 ]

Then solving the matrices:

for 2 + i : [ 1 - i , 1 , 0 ; -2 , -1 - i , 0] and using the fact that the equations are multiples of each other, I cleared out the second row, which left me with (1-i)x_1 + x_2 = 0. I let x_2 = t, solved for x_1 using this free parameter, rationalized my denominator, etc. I did the same for my other eval.

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: Suppose that i + 1 is an eigenvalue of a matrix A and [-1 + i, i ] is a corresponding eigenvector. Find a fundamental set of real solutions to the equation y ' = A y.

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Your solution:

y = c_1[-e^tcost e^tsint ; -e^tsint] + c_2[ e^tcost - e^tsint ; e^tcost]

confidence rating #$&*:

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Solve the equation

y ' = [0, -9; 1, 0] y

with initial condition

y(0) = [6, 2].

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Your solution:

y = 2[ 3sin(-3t) ; cos(-3t) ] -2[ -3cos(-3t) ; sin(-3t) ]

Solving for my evals, I found lambas = +/-3i

I used lambda= -3i to find the evector: [-3i ; 1]

I then set up (cos(-3t) + isin(-3t) * [-3i ; 1] to find the real and imaginary parts of the solution, factored the out, and found c_1 = 2 and c_2 = -2 when I imposed the initial condition.

confidence rating #$&*:

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Find all values of mu such that any fundamental set [ y_1, y_2 ] of the system

y ' = [1, 3; mu, -2] y

have the property that the limit of the expression (y_1(t))^2 + (y_2(t))^2, as t -< infinity, is zero.

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Your solution: none

confidence rating #$&*:

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I honestly didn't know where to start with this one. I got as far as finding my evals to be -1/2 +/- sqrt(-7 - 12mu)/2

and I've convinced myself that I must not be approaching the problem correctly.

@&

If sqrt(-7-12 mu) / 2 is imaginary, then the solutions will be of the form

e^(-t/2) * (bounded linear combination of sines and cosines)

and will therefore converge as t -> infinity.

This will be the case for -7 - 12 mu < 0, so that mu > -12/7.

If mu <= -12/7, then sqrt(-7-12 mu) / 2 is real and the question will be whether -1/2 + sqrt(-7-12 mu) is positive. If so, the solution won't converge at infinity.

*@

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Self-critique rating: 3

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Question: A particle moves in an unspecified force field in such a way that its position vector r(t) = x(t) i + y(t) j and the corresponding velocity vector v(t) = r ' (t) satisfy the equation

v ' = 2 k X v

Write this condition as a system

v ' = A v,

with v = [v_x; v_y].

If the particle starts at position r(0) = 2 i + j, v(0) = i + 2 j, find its position at t = 3 pi / 2.

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Your solution:

[v_x ; v_y ]' = [0 , -2 ; 2 , 0 ][v_x ; v_y]

v = [-2sin2t , cos2t ; 2cos2t , sin2t]

r = [cos2t , .5sin2t ; sin2t , -.5cos2t] + C

r(3pi/2) = 5/2j

I came out with evals of +/- 2i and corresponding evecs of [i ; 1] and [-i ; 1]

I then used these to come up with a real valued solution

confidence rating #$&*:

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): I feel ok about my solution up to where I had to integrate v. I am confident I integrated correctly, but I didn't really know how to factor the initial condition into that.

@&

When you integrate you get integration constants, which allow you to satisfy the initial conditions.

*@

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Self-critique rating: 3"

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