Assignment 27

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course Mth 279

8/5

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]

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

= -sinh + sinh + e^t - e^-t

W(t) = e^t - e^-t

so linearly independent

Definitely not confident at all on taking the determinate that I just did??

Not sure if the matrix is suppose to be [e^t,1; e^-t,1; sinh(t),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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The determinant doesn't help in this case. You can only take the determinant of a square matrix, and the matrix you get from the three solutions is not square.

Since sinh(t) = (e^t - e^-t) / 2, it follows that y_1 - y_2 = 2 y_3.

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],

which is a linear combination of the vectors y_1, y_2 and y_3, with not all constants equal to zero, but with the linear combination equal to zero. This is the definition of linear dependence.

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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; sin^2, 2-2cos^2, -2; 0, 0,1]

W(t) = 1((2-2cos^2(t)(1) - (-2*0)) - 0((sin^2(t)(1) - (-2*0)) + 1((sin^2(t)(0) - (2-2cos^2(t)(0))

W(t) = 2-2cos^2(t) -2

W(t) = -2cos^2(t)

Linearly independent

???Not sure what I am doing at all???

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You misplaced a term, but you're doing the right things.

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

= 2 - 2 cos^2(t) - 2 sin^2(t)

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

= 2 - 2(1) = 0.

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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: 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]

W(t) = det(psi) = t^2(1)-2t(0) = t^2

yes fundamental set of solutions

psi’(t) = [2t, 2; 0,0]

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

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

P(t) = [2/t, -4; 0,0]

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Very good. However

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

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

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

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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:

a) y_1 = [e^t;e^t] y_2= [e^-t; -e^-t]

b) y’ = P(t)y

P(t) = [0,1; 1,0]

y’_1 = [0,1; 1,0][e^t;e^t] = [ -e^t; e^t]

y’_2 = [0,1; 1,0][e^-t; -e^-t] = [-e^-t; -e^-t]

I have no idea how to answer and solve the rest of the questions??? I feel that this last chapter and assignments are all over the place. I am just not comprehending what I need to do and how to do each of these questions???

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y1 = [e^t; e^-t] and

y2 = [e^t, -e^t].

y = [y1; y2], so the equation y ' = P(t) y is

[y1'; y2'] = [0,1;1,0] * [y1, y2].

Written out this system is

y1 ' = y2

y2 ' = y1

The column vector y1 = [e^t; e^-t] is a solution to y ' = P(t) y since

[e^t; e^-t] ' = [e^t; -e^(-t) ] (note that this is equal to y2)

and P(t) y1 = [0,1; 1,0] * [e^t, e^-t; e^t, -e^(-t)] = [e^t; -e^(-t) ].

A similar calculation will show that y2 is also a solution.

The Wronskian of the solution set is easily seen to be -2, showing that [ y1, y2 ] is a fundamental set.

psi ' (t) = [e^t, -e^-t; e^t, e^(-t)] and is easily shown to be equal to P(t) psi(t).

With

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

we find the desired matrix C by solving

psi_hat = psi * C

for C, obtaining

C = psi^-1 * psi_hat.

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

The matrix product psi^-1 * psi_hat is easily calculated to obtain the constant matrix C.

If C is invertible then psi_hat is a fundamental matrix for this system.

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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: 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:

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

yes fundamental set

2) How do you to let t=0???Or why do you let t=0??

psi(0) = [e^0, e^-2(0); 0, e^(-2*0)] = [1, 1; 0, 1]

psi_hat(0)= I = psi(t)*C

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

psi_hat * psi^-1 = C

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

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

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Good.

Check:

psi(0) * C = psi_hat(0)

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

Which checks out.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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