Query_27

#$&*

course MTH-279

08/06 around 3:00 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:

p(lambda) = det [A - lambda I]

I = identity matrix [1, 0 ; 0, 1] so, lambda I ]= [lambda, 0 ; 0, lambda]

det [3, 1 ; -2, 1] - [lambda, 0 ; 0, lambda]

= det [ 3 - lambda, 1 - 0 ; -2 - 0, 1 - lambda]

=det[3 -lambda, 1; -2, 1-lambda]

=[(3-lambda * 1-lambda)] - [(-2 * 1)]

= lambda^2 - 4lambda + 5

use quadratic formula to find lambda = 2 +/- i

---

lambda = 2 + i

(A - lambda I) * vector x1 = 0

[1-I, 1 ; -2, -1-i] * [x1; x2] = [0;0]

(1-i)x1 + x2 = 0

-2x1 + (-1 - i)x2 = 0

-> 2x1 + (1 + i)x2 = 0

---

(1-i)x1 + x2 = 0

2x1 + (1+i)x2 = 0 - plug in -i

---

(1-i)-I + x2 = 0

-i + i^2 = -x2

-i - i^2 = x2

I + 1 = x2

---

2x1 + (1+i)(i+1) = 0

2x1 + (i^2 + 2i + 1) = 0

2x1 + 2i = 0

2i = -2x1

x1 = -i

so, x1 vector = [-I; i + 1]

--now repeat the process for lambda 2 to find x2 vector—

we find that x2 vector = [i; -i + 1]

Confidence rating: 3

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Given Solution:

[see my note at end for correct (?) solution ]

[tilson]

matrix is [3, 1; -2, 1]

Assume the solution will take the form y(t) = e^(λt)x, so y’ = λe^(λt)x,

λe^(λt)x = Ae^(λt)x,

e^(λt)(λx) = e^(λt)(Ax),

λx = Ax,

(A-λI)x = 0, x ≠ 0.

det([3-λ, 1; -2, 1-λ]) = 0

λ^2 - 4λ + 5 = 0

λ = 2 ± i

For λ = 2 + i:

[3-2-i, 1; -2, 1-2-i] = [1-i, 1; -2, -1-i]. Using row reduction, we get [2, 1+i; 0,0]. This can be interpreted as 2x + (1+i)y = 0. Putting this into vector form, we get: v_1=[-1-i; 2].

For λ = 2-i:

[3-2+i, 1; -2, 1-2+i] = [i+i, 1; -2, -1+i]. Using row reduction, we get [2, 1-i; 0, 0]. This can be interpreted as 2x + (1-i)y = 0. Putting this into vector form, we get: v_2 = [-1+i; 2].

[white]

Phi = [-4 +5e^t , 4 - 4e^t ; -5+5e^t , 5-4e^t]

y(3) = [e^2 + 1 ; e^2]

using the fact that e^(tA) = T*lam*T_inv

I found my evals to be 0 and 1 and my evecs to be [4 ; 5] and [1 ; 1]

so T = [4 , 1 ; 5 , 1] T_inv = [-1 , 1 ; 5 , -4]

I then computed [4 , 1 ; 5 , 1]*[1 , 0 ; 0 , e^t]*[-1 , 1 ; 5 , -4] = [-4 +5e^t , 4 - 4e^t ; -5+5e^t , 5-4e^t]

to find y(3), I set up [-4 +5e^(t-s) , 4 - 4e^(t-s) ; -5+5e^(t-s) , 5-4e^(t-s)] = [y_t1 - y_s1 ; y_t2 - y_s2]

or [-4 +5e^2 , 4 - 4e^2 ; -5+5e^2 , 5-4e^2] = [ y_t1 - 1 ; y_t2 + 0]

and solved for y.

[my note]

@&

The characteristic equation is

det( [5-lambda, -4; 5, -4-lambda] )

= (5 - lambda)(-4-lambda) - (5)(-4)

= lambda^2 - lambda

The eigenvalues are 0 and 1.

Thus for lambda = 0

5 y_1 - 4 y_2 = 0 and

y_2 = 5/4 y_1

so an eigenvector is

[1; 5/4] .

If we want integer coefficients we can use

[4;5].

The solution vector corresponding to eigenvalue 0 is therefore

e^(0 t) [4; 5] = [4, 5].

For lambda = 1 the eigenvector [1; 1] works, leading to solution

e^t [1; 1].

The corresponding phi(t) matrix is

phi(t) = [4, e^t ; 5, e^t].

Checking that this matrix solves the equation we have

phi ' (t) = [0, e^t; 0, e^t]

and

[5, -4; 5, -4] * phi(t) = [0, e^t; 0, e^t]

so that phi(t) is a solution to the system.

Now the general solution y(t) is

y(t) = phi(t) * [c_1; c_2],

which must be equal to [1, 0]

The inverse of phi(t) is

[-e^t, e^t; 5, -4] / e^t = [-1, 1; 5 e^(-t), -4 e^(-t) ]

When t = 0 this becomes just

[-1, 1; 5, -4]

so that the solution for the c matrix is

[c_1; c_2] = [-1, 1; 5 , -4 ] * [1; 0] = [-1, 5].

Thus our desired solution is

[4, e^t; 5, e^t] * [-1; 5] = [-4 + 5 e^t; -5 + 5 e^t ]

It is easy to check that this is a solution.

Evaluated at t = 0 it does indeed give us [1; 0].

Its derivative is

[5 e^t; 5 e^t]

and its product with the matrix [5, -4; 5, -4] is equal to this result.

*@

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Self-critique (if necessary): OK

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:

Confidence rating:

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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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:

remember y’ = Alambda

p(lambda) = det [A - lambda I] = 0

--follow the steps similar to those in problem one—

we find lambda = +/- 3i

--now solve for the eigenvectors—

using [A - lambda1 I] * x1vector = [0;0]

and [-3i, -9 ; 1 , -3i] * [x1; x2] = [0;0] for reference

x2 = 1/3 and x1 = i, so x1vector = [i; 1/3]

same method shows that x2vector = [-i; 1/3]

y1(t) = e ^ 3it [ i 1/3]

y1(t) = [ie^3it ; 1/3 e^3it]

---

y2(t) = e^-3it [-i; 1/3]

y2(t) = [-ie^-3it; 1/3 e^-3it]

---

y(t) = c1y1(t) + c2y2(t)

= [ie^3it , -ie^-3it ; 1/3e^3it , 1/3 e^-3it] * [c1; c2]

---

w(t) = det [above matrix w/ zero plugged in]

=det[i, -i ; 1/3, 1/3]

=(i * 1/3) - (1/3 * -i)

= 1/3i + 1/3i

= 2/3i

w(t) does not equal zero

---

y(0) = [i , -i ; 1/3, 1/3] * [c1; c2] = [6;2]

distribute and rearrange

[c1; c2] = [3i + 3; -3i + 3]

--

y(t) = 3i + 3 [ie^3it; 1/3 e^3it] -3i + 3 [ -ie^-3it ; 1/3 e^-3it]

Confidence rating: 2,3

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Given Solution:

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Self-critique (if necessary): OK

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

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

[tilson]

lim(e^(-(t^2))) as t ∞ = 0.

As long as the solution takes a form of e^(-(t^2)), the limit will be 0. In order to get a solution that will take that form when squared, we’re looking for complex eigenvalues, since (±i)^2 = -1. It doesn’t really matter if we have the positive or negative eigenvalue, as long as it is complex.

We find the eigenvalues as normal,

det([1-λ, 3; µ, -2-λ]) = 0,

λ^2 + λ - (3µ + 2) = 0.

Using the quadratic formula, we get:

-1/2 ± sqrt(9+12µ)/2. We know that we need the sqrt()/2 term to equal a multiple of i. Knowing this, we set up an equality: sqrt(9+12µ)/2 = ni.

Solving, we get µ = -n^2/3 - 3/4. This means that n can be any real value. In that regard, µ can take any value such that -12µ is greater than 9.

@&

9 + 12 mu < 0 if mu < 9/12 = 3/4; this would result in imaginary solutions

If

sqrt(9 + 12 mu) / 2 > 1/2

then one eigenvalue would be positive, leading to a solution that doesn't converge at infinity.

*@

note also following

[white]

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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Your solution:

Confidence rating:

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Given Solution:

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Self-critique (if necessary):

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:

Confidence rating:

\Given Solution:

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].

I’m a bit stumped by this problem. The X in v’ = 2kXv is throwing me off. I’m not sure if that’s supposed to indicate the cross product, as in “2k crossed with v”, or if X is intended to be a separate matrix, or if 2k is even supposed to represent 2 units in the k direction. Since you’ve used i and j to denote directions in r(t), I’m assuming that k is the k direction. But the X is still throwing me off. I’ll just assume that it’s “2k crossed with v.”

@&

That is the correct assumption.

*@

[0,0,2]x[v_x, v_y, v_z] = [-2v_y, 2v_x],

v’ = [0, -2; 2, 0][v_x; v_y],

@&

Good, but

[v_x ' ; v_y '] = [0, -2; 2,0] [v_x; v_y] would be more explicit

v ' = [0, -2; 2, 0] v is another alternative.

@

det([-λ, -2; 2, -λ]) = 0,

λ^2 + 4 = 0,

λ = ±2i.

For λ = -2i:

[2i, -2; 2, 2i]. Using row reduction, we get: [1, i; 0, 0]. We interpret this as x+iy=0. We translate this

into vector form as v_1 = [1; i].

@&

fine, but maybe [v_x; v_y] = [1; i] to emphasize the details

@

v(t) = e^(-2it)[1;i] = (cos(2t) - i*sin(2t))[1;i] = u + iv = [cos(2t); sin(2t)] + i[-sin(2t); cos(2t)]

v(t) = c_1[cos(2t); sin(2t)] + c_2[-sin(2t);cos(2t)]

v(0) = [c_1cos(0) - c_2sin(0); c_1sin(2t) + c_2cos(2t)] = [1;2],

c_1 = 1, c_2 = 2.

r(t) = ʃv(t)dt = [cos(2t) + sin(2t)/2 + c_3; sin(2t) - cos(2t)/2 + c_4].

r(0) = [1+c_3; c_4 - ˝] = [2; 1].

c_3 = 1, c_4 = 3/2.

r(t) = [cos(2t) + sin(2t)/2 + 1; sin(2t) - cos(2t)/2 + 3/2]

r(3π/2) = [0; 2].

The particle’s position is 2j.

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Self-critique (if necessary):

Self-critique rating:

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