Assignment 29

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course Mth 279

8/5

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:

1)det(A-lI)= 0

det([3,1;-2,1]-l[1,0;0,1])

det(3-l, 1; -2, 1-l) = ((3-l)(1-l))-(1*-2))

0 = l^2 - 4l + 5

l= 2+-i

eigenvalues

l_1 = 2+i

l_2 = 2-i

2) l_1 = 2+i

[1-i, 1; -2, -1-i][k1;k2]=0

rref

[2, 1+i; 0,0][k1;k2] = 0

2k_1 + (1+i)k_2 = 0

k_1 = -(1+i)/2 k_2

letk_2 = 2

k_1 = [-1+i;2] eigenvector

l_2 =2-i

[1+i, 1; -2, -1+i][k1;k2] = -

rref

[2, 1-i; 0, 0] [k1;k2] = 0

2k_1 + (1-i)k_2 = 0

k_1 = -(1-i)/2 k_2

k_2 = -2

k_2 = [1-i; -2] eigenvector

Could you solve for k_1 and k_2 without taking rref???

Can you choose k_2 to be anything for ex. k_2 = 1 instead of 2 and still get a correct answer???

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Yes. However, either 2 or -2 would be good choices.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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(t) = e^(lt)k

y(t) = e^((i+1)t)[-1+i;i]

=e^(it)e^t[-1+i;i]

By Euler’s formula y(t) becomes

y(t) = e^t(cos(t)+ isin(t))[-1+i;i]

= [-cos(t) - isin(t) + icos(t) -sin(t); icos(t)-sin(t)]

separating real and imaginary solution

y(t) =e^t[-cos(t)-sin(t);-sin(t)]+e^t[-sin(t)+cos(t);cos(t)]i

y_1(t) = e^t [-cos(t)-sin(t);-sin(t)]

y_2(t) = e^t [-sin(t)+cos(t);cos(t)]

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Given Solution:

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Self-critique (if necessary):

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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:

1) det(A-lI)= 0

det(-l, -9; 1, -l] = 0

l^2+ 9= 0

l= +-3i

2)l_1 = 3i

[-3i,-9;1,-3i][k1;k2] = 0

rref

[-3i, -9; 0,0] [k1;k2]=0

-3ik_1 -9k_2 = 0

k_1 = -3/i k_2

let k_2 = i

k_1 = [-3;i]

3) l_2 = -3i

[3i, -9; 1, 3i][k1;k2] = 0

rref

[3i, -9;0,0][k1;k2]=0

3ik_1 - 9k_2 = 0

k_1 =3/i k_2

let k_2 = i

k_2 = [3;i]

4) y(t) = e^(3it)[3;i]

use Eulers

y(t) = (cos(3t) + isin(3t))[3;i]

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

y_1 = [3cos(3t);-sin(3t)]

y_2 = [3sin(3t); cos(3t)]

y(t) = C_1y_1 + C_2y_2

y(0) = [6;2]

C_1[3cos(0);-sin(0)] + C_2[3sin(0);cos(0)] = [6;2]

3C_1=6

C_1 = 2

C_2 = 2

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

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

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

has 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:

det([1,3;u, -2] - l[1,0;0,1])

det(1-l, 3; u, -2-l)

= l^2+l -2-3u

use quadratic formula to find l

l = -1+-sqrt(9+12u)/2

eigenvalues will be real when

9+12u >0

u>-3/4

eigenvalues will be complex when

9+12u <0

u <-3/4

Unsure what I need to do once eigenvalues are found to answer question about (y_1(t))^2 + (y_2(t))^2, as t -> infinity, is zero.???

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If eigenvalues are complex then the real part being -1/2, the solution will be of the form

e^(-1/2) * (sum of sines and cosines)

and will approach zero as t -> infinity.

If sqrt(9 + 12 mu) < 1 the result will be real solutions with negative multiples of t as exponents, and will again approach zero.

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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: 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].

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Your solution:

No idea how to go about this problem???

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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