Query 25

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course MTH 279

4/15 4

Query 25 Differential Equations*********************************************

Question: Determine whether the set of solutions {y_1, y_2, y_3} is linearly independent, where

y_1 = [ e^t, 1]

y_2 = [ e^(-t), 1]

y_3 = [ sinh(t), 0]

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Your solution:

Linearly independent.

Since the determinant of a 2x3 can't be computed, the only other thing I could think of to look for is if the equations are scalar multiples, which they do not appear to be.

@&

The vectors are linearly dependent.

The reason is that sinh(t) is a linear combination of e^t and e^(-t).

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I do not know if there was another test for linear independence I wasn't considering.

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Self-critique rating: 3

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Question: Determine whether the set of solutions {y_1, y_2, y_3} is linearly independent, where

y_1 = [ 1, sin^2(t), 0]

y_2 = [ 0, 2 - 2 cos^2(t), -2]

y_3 = [ 1, 0, 1]

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Your solution:

Yes. det = 2-2cos^2t, which is nonzero on -inf < t < inf

@&

The determinant is

1 * det[2 - 2 cos^2 (t), -2; 0, 1] - sin^2(t) det[0, -2; 1, 1] + 0 det[0, 2 - 2 cos^2(t); 1, 0]

= 2 - 2 cos^2(t) - 2 sin^2(t) = 2 - 2 ( cos^2(t) + sin^2(t)) = 2 - 2 = 0.

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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: Determine whether there is a matrix P(t) such that

y_1 = [ t^2, 0 ]

y_2 = [ 2t, 1 ]

is a fundamental set of solutions to the equation

y ' = P(t) y.

If so, find such a matrix P(t).

Hint: The matrix psi(t) = [y_1, y_2 ] = [ t^2, 2 t; 0, 1 ] would need to satisfy psi ' (t) = P(t) psi(t).

In standard notation we could write this as follows:

satisfies

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Your solution:

P = [ 2/t , 2/t^2 ; -2 , 0]

@&

I don't believe psi ' (t) = P(t) psi(t) for this particular P matrix.

You don't show your steps so I can't tell how you obtained your matrix. However multiplying psi ' (t) on the right by psi^-1(t) would give you a P matrix that works.

Consider

P = [ 2t, -2; 0, 0]

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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: If the matrix psi(t) = [y_1, y_2] = [e^t, e^(-t); e^t, - e^(-t)]:

What are the vector functions y_1 and y_2?

Write out the system of two differential equations represented by the equation y ' = P(t) y with P(t) = [0, 1; 1, 0].

Show that y_1 and y_2 are both solutions of the equation y ' = P(t) y with P(t) = [0, 1; 1, 0].

Show that { y_1 , y_2} is a fundamental set for this equation.

Show that the matrix psi(t) is a solution of the matrix equation psi ' = P(t) psi.

Show that the matrix psi(t) is a fundamental matrix for the linear system of equations.

Let psi_hat(t) = [ 2 e^t - e^(-t), e^t + 3 e^(-t); 2 e^t + e^(-t), e^t - 3e^(-t) ].

Find a constant matrix C such that psi_hat(t) = psi(t) * C.

Based on your matrix C, is psi_hat(t) a solution matrix for the system?

Based on your matrix C, is psi_hat(t) a fundamental matrix for the system?

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Your solution:

y_1 = [e^t ; e^t]

y_2 = [e^-t ; -e^-t]

[e^t ; e^t] = [0, 1; 1, 0] * [e^t ; e^t]

[-e^-t ; e^-t] = [0, 1; 1, 0] * [e^-t ; -e^-t]

doing matrix multiplication on the right hand sides of these two equations, we do see [e^t ; e^t]= [e^t ; e^t] and [-e^-t ; e^-t] = [-e^-t ; e^-t]

psi = [ e^t , e^-t ; e^t , -e^-t]

det(psi) = -2, so yes. y_1 and y_2 are a fundamental set.

[ e^t , -e^-t ; e^t , e^-t] = [0, 1; 1, 0]*[ e^t , e^-t ; e^t , -e^-t]

yes, psi is a solution.

C = [-4 , -2 ; 2 , -6]

since C is invertible, psi_hat is a solution matrix. it is also a fundamental matrix.

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating:

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Question: Given the system

y ' = [ 1, 1; 0, -2 ] y

verify that

psi(t) = [ e^t, e^(-2 t); 0, e^(-2 t) ]

is a fundamental matrix for the system.

Find a matrix C such that

psi_hat(t) = psi(t) * C

is a solution matrix satisfying initial condition psi_hat(0) = I, where I is the identity matrix.

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Your solution:

psi is not a fundamental matrix.

[ e^t, -2e^(-2 t); 0, -2e^(-2 t) ] is not equal to [1 , 1 ; 0 , -2]*[ e^t, e^(-2 t); 0, e^(-2 t) ] = [e^t , 2e^(-2t) ; 0 , -2e^(-2t)]

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary): I found it odd that the right hand side differed from the left by a negative sign on only the 1,2 term, but I could not find where I made an error in my calculations.

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Self-critique rating:3

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