Query 25

#$&*

course MTH 279

5/4 11:24 am

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: 

psi_1 = [e^t, e^-t; 1, 1] --> W = e^t - e^-t

psi_2 = [e^t, sinh t; 1, 0]--> W = 0-sinh t

psi_3 = [e^-t, sinh t; 1, 0] --> W = 0- sinh t

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Given Solution: 

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Self-critique (if necessary):

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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: 

 psi = [1, 0, 1; sin^2 t, 2-2cos^2 t, 0; 0, -2, 1]

W = det psi = 2-2(cos^2 t + sin^2 t)= 0

so not linearly independent

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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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: 

psi = [ t^2, 2t; 0, 1]

W = det psi = t^2 ≠ 0 so Fundamental

psi' = [2t, 2; 0, 0]

psi^-1 = 1/t^2*[1, -2t; 0, t^2] = [1/t^2, -2/t; 0, 1/t]

psi'*psi^-1 = P(t) = [2t, 2; 0, 0]* [1/t^2, -2/t; 0, 1/t]= [2/t, -4; 0, 0]

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:  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]

Here's trouble for me in writing out the individual eqs in the system

y_1' = P*y_2

y_2' = P*y_1

this is based on section 6.3 but is clearly not right

but y_1' = P*y_1 and y_2' = P*y_2 does work so that y_1 and y_2 are solutions to that system but not to the one I came up with

@& y_1 ' = [ e^t; - e^-t]

P(t) y = [y_2, y_1]

so the system is

y_1 ' = y_2

y_2 ' = y_1

P(t) gives you this system, but doesn't appear in the system, having already done its work byt giving you the system.

*@

W({ y_1 , y_2}) = -2

W(y_1) =

psi ' = [e^t, -e^-t; e^t, e^-t]

psi^-1 = -1/2 * [-e^-t, -e^-t; -e^-t, e^t]= [1/2e^-t, 1/2e^-t; 1/2e^-t, -1/2e^t]

psi' *psi^-1 = [e^t, -e^-t; e^t, e^-t]*[1/2e^-t, 1/2e^-t; 1/2e^-t, -1/2e^t] = [0,1; 1,0] = P

W= det psi = (-e^-t * e^t) - (e^-t* e^t) = -2 ≠ 0 so Fundamental

C = [2, 1; -1, 3]

det C = 7 so fundamental solution set

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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Self-critique rating:

@& Good. Check my note on that one step. I think you used P(t) twice instead of once.*@

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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: 

W(psi) = det [ e^t, e^(-2 t); 0, e^(-2 t) ] = e^-t so Fundamental

psi(0)= [1,1; 0, 1] psi^-1 = [1, -1; 0, 1] --> C = psi^-1(0) = [1, -1; 0, 1]

  

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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#*&!

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