Query_25

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course MTH-279

08/03 around 11:20 PM

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(t) = [e^t, e^-t, sinh(t); 1, 1, 0]

= [e^-t, sinh(t); 1, 0] - [e^t, sinh(t) ; 1,0] + [e^t, e^-t ; 1, 1]

=[(e^-t * 0) - (1 * sinh(t)] - [(e^t * 0) - [1* sinh(t)] + [(e^t * 1) - (1 * e^-t)]

= [(0) - sinh(t)] - [(0) - sinh(t)] + [(e^t - e^-t)]

= e^t - e^-t -> 2sinh(t)

yes, the solutions are linearly independent and fundamental set

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Since sinh(t) = (e^t - e^-t) / 2, it follows that y_1 - y_2 = 2 y_3, which is what you have shown.

So the solutions are not linearly independent.

Formally, if c_1 = 1, c_2 = 1 and c_3 = -2, we have

c_1 y_1 + c_2 y_2 - 2 c_3 y_3 = [0; 0]

*@

Confidence rating: 2

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Given Solution:

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Self-critique (if necessary): OK

Self-critique rating: OK

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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(t) = [1, 0, 1; sin2t, 2-2cos^2(t), 0 ; 0, -2, 1]

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

1 * [(2-scos^2(t)*1) - (-2*0)] + 1 * [(sin^2(t) * -2) - (0 * 2 - 2cos^2(t))]

= 1 * [2 - 2cos^2(t)] + 1* [-2sin^2(t)]

---

= 2 - 2cos^2(t) - 2 sin^2(t) -> -2(cos^2(t) + sin^2(t) - 1) -> -2 (1-1)

= 0, not linearly independent

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: 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) = [t^2, 2t; 0, 1]

note: inverse of 2x2 matrix is 1/det(A) [d, -b ; -c, a]

---

y’ = p(t)y

psi’(t) = p(t)psi(t)

p(t) = psi’(t) * psi^-1(t)

psi’(t) = [2t, 2; 0, 0]

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

p(t) = psi’(t) * psi^-1(t) = [2t, 2; 0, 0] * [-1/t^2, -2t/t; 0, 1]

= [2/t + 0, -4 + 2 ; 0 + 0, 0 + 0] = [2/t, -2; 0, 0] -> p(t)

Confidence rating: 3

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Given Solution:

Self-critique (if necessary): OK

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:

Confidence rating: 2

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Given Solution:

Did this on paper, quite a bit of work, but got answers and concepts.

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Self-critique (if necessary): OK

Self-critique rating: OK

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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(t) = (e^t * e^-2t) - (0 - e^-2t) -> e^-t - 0

= e^-t, so is a fundamental set

---

psi^(0) = psi(0) + C

I = [1, 0 ; 0, 1] = [1, 1; 0, 1] * C

---rearrange

C = psi^(0) + psi^-1(0)

C = [1, 0; 0, 1] * 1/1 [1, -1; 0, 1]

C = [1, 0; 0, 1] * [1, -1; 0, 1] = [1+0, -1+0 ; 0+0, 0+1]

= [1, -1; 0, 1] = C

Confidence rating: 3

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Given Solution:

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Self-critique (if necessary):OK

Self-critique rating: OK

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