query 24

#$&*

course Mth 279

4/30 3

Query 24 Differential Equations*********************************************

Question: Verify Abel's Theorem in the interval (-infinity, infinity) for

y ' = [ 6, 5; -7, -6] * y

whose solutions are

y_1 = [ 5 e^-t; -7 e^-t ]

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

with t_0 = -1

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Your solution:

For y_1 = [ 5 e^-t; -7 e^-t ] and y_2 = [ e^t; - e^t ] when we combine we get `psi

`psi(t) = [5 e^-t, e^t; -7 e^-t, - e^t], calculating the determinant

det[`psi(t)] = [5e^(-t)*-e^(t)] - [-7e^(t)*e^(t)] = 2(for any value of t)

W(t) = det[`psi(t)] = 2

Able’s Theorem states that if {y_1, y_2,…..y_n} is a set of solutions of y’ = P(t)y, a

Knowing the Wronskian of the solutions. Then, W(t) satisfies the scalar differential equation

W’(t) = tr[P(t)]*W(t)

Moreover, if t_0 is any point in a

W(t) = W(t_0)*e^(Int(tr[P(s)] ds,t_0,t))

tr[P(s)] = [6 5; -7,-6] = 0, so the Int( tr[P(s)] ds,t_0,t) = 0(t - t_0) = 0

So

W(t) = W(t_0)*e^( Int(tr[P(s)] ds,t_0,t)), plugging in our values

2 = 2*e^(0) = 2*1 = 2, which checks out

?????So this is the process, but not completely sure what’s going on here. Not even sure I got the process completely right. If you can please tell me what I’m missing??????????

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

@& You got the process.

The trace of the matrix is zero, so the Wronskian is constant (since e^0 = 1).*@

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Question: y ' = A y, with solutions

y_1 = [5; 1]

y_2 = [2 e^(3 t), e^(3 t) ]

Verify that this constitutes a fundamental set.

Find Tr(A).

Show that

psi(t) = [y_1, y_2]

satisfies

psi ' = A * psi

Find A by finding psi ' * psi^-1

Is the result consistent with your result for the trace of A?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For y_1 = [5; 1] and y_2 = [2 e^(3 t), e^(3 t) ] when we combine we get `psi

`psi(t) = [5, 2e^(3t); 1, e^(3 t)], calculating the determinant

det[`psi(t)] = 3e^(3t),

Determinant was non-zero verifying y_1 and y_2 constitutes a fundamental set.

????????

A=[5, 2e^(3t); 1 , e^(3t)]

Tr[A] = 5+e^(3t)

????????

psi(t) = [y_1, y_2] = [5, 2e^(3t); 1 , e^(3t)]

????I’m confused what the question wants me to do??????

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

@& psi ' (t) = [ 0, 6 e^(3 t); 0, 3 e^(3 t) ]

W(0) = 3, so

W(t) = 3 e^(3 t) = W(0) e^(3 t).

W(t) = W(0) e^integral(Tr(A) dt) ,

so Tr(A) must be 3.

A psi = [ A y_1, A y_2 ]

where A y_1 and A y_2 are column vectors y_1 'and y_2 '.

psi^-1 = 1 / (3 e^(3 t) * [e^(3 t), -2 e^(3 t); -1, 5 ]

so

A = psi ' psi^-1 = [-2, 10; -1, 5].

You can verify that psi ' found above is equal to A psi for your psi matrix.

*@

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Question:

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Your solution:

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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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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