Query 26

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course MTH 279

4/19 7pmI apologize for getting these queries 26 and 27 turned in so late. I could have gotten them in earlier, but not with the same quality. I have struggled with the matrix math a bit, so I decided to finish off my linear algebra class before I finished these queries.

Query 26 Differential Equations

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Question: Find the solutions to y ' = A y when

A = [ 5, 3; -4, -3 ]

and

y(1) = [2; 0].

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Your solution:

lambda_1 = 3 and lambda_2 = -1

Found using det( 5 - lambda , 3 ; -4 , -3 - lambda) = 0

Solving (A - lambda*I) = [ 0 ; 0]

gives eigenvectors of [-1.5 ; 1] for lambda_1 = 3 and [-.5 ; 1] for lambda = -1

So we have the general solution:

y= c_1[-1.5e^3t ; e^3t] + c_2[-.5e^-t ; e^-t]

y= e^-3[-1.5e^3t ; e^3t] + e[-.5e^-t ; e^-t]

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Find the solutions to y ' = A y when

A = [ 4,2,0; 0,1,3; 0,0, -2 ]

and

y(0) = [-1;0;3].

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Your solution:

y=3[ 1/3e^-2t ; -e^-2t ; e^-2t] + 3[ -2/3e^t ; e^t ; 0]

det(A - lambda*I) ' det[4 - lambda , 2 , 0 ; - , 1- lambds , 3 ; 0 , 0 , -2 - lambda]

gives us lambda_1 = -2, lambda_2 = 1, lambda_3 = 4

eigenvector for lambda_1 = (A - -2*I)x=[ 0 ; 0 ; 0]

gives x=[ 1/3 ; -1 ; 1]

eigenvector for lambda_2 = (A - 1*I)x=[ 0 ; 0 ; 0]

gives x= [-2/3 ; 1 ; 0]

@&

It's generally easier to work with whole numbers. Rather than

[ 1/3 ; -1 ; 1] and [-2/3 ; 1 ; 0]

we can use any nonzero multiple of either, so that

[1, -3, 3] and [-2, 3, 0]

might be simpler choices.

However this would be just a choice of convenience; there is nothing invalid about using the fractions.

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eigenvector for lambda_3 = (A - 4*I)x=[ 0 ; 0 ; 0]

gives x = [1 ; 0 ; 0]

This third eigenvector I'm not too sure about the row reduced matrix I came out with was [ 0 1 0 ; 0 0 1 ; 0 0 0], which I interpreted as x_1 = t , x_2 = 0 and x_3 = 0. That seemed a little odd to me.

@&

That is right.

The matrix equation tells you that x2 = 0 and x3 = 0.

x1 could be anything, so x1 = 1 is valid, and is the simplest choice.

*@

I got c_1 = 3, c_2 = 3, and c_3 = 0

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Three tanks, each full of a solution and all of equal volume V, are each connected to each of the others by two pipes. Each tank also has a third pipe through which pure water flows into it, and a fourth through which water exits.

The flow rate r through every pipe is the same.

Write the system of equations that relates the quantities Q_1, Q_2 and Q_3 representing the amount of solute in each tank as a function of time.

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Your solution:

change in conc. wrt t = rate of inflow - rate of outflow

and basically we have 3 outflows in each pipe (each pipe is connected to two others and another that water exists out of) and 2 inflows, one from each of the other tanks. I'm setting this up assuming that the pipes that connect the tanks are simultaneously in and outflow pipes....which may not be correct. This gives us a system of 3 equations...

Q_1' = -3(r/V)Q_1 + (r/V)Q_2 + (r/V)Q_3

Q_2' = (r/V)Q_1 - 3(r/V)Q_2 + (r/V)Q_3

Q_3' = (r/V)Q_1 + (r/V)Q_2 - 3(r/V)Q_3

which gives us the matrix equation:

[Q_1 ; Q_2 ; Q_3 ]' = r/V [ -3 , 1 , 1 ; 1 -3 , 1 ; 1 , 1 , -3] * [ Q_1 ; Q_2 ; Q_3]

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK"

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

@&

No problem with being a little bit late. It makes sense to finish the linear algebra course, since the methods used in this part of our course are so dependent on linear algebra.

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