Assignment 25

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course Mth 279

8/5

Query 23 Differential Equations*********************************************

Question: What is the largest interval over which the solution to the system

(t + 2) y_1 ' = 3 t y_1 + 5 y_2

(t - 2) y_2 ' = 2 y_1 + 4 y_2

with initial conditions

y_1(1) = 0

y_2(1) = 2

is defined?

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Your solution:

standard form: y’_1 = 3ty1/(t+2) + 5y2/(t+2)

y’_2 = 2y1/(t-2) + 4y2/(t-2)

find where P(t) and g(t) are discontinuous

3t/(t+2)

t=-2

2/(t-2) t=2

so largest intveral (-2,2)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: The equation

y ' = [ y_2; y_3; -2 y_1 + 4 y_3 + e^(3 t) ]

with initial condition

y(0) = [1; -2; 3]

represents a higher-order equation of form

y[n] + a_(n-1) * y[n-1] + ... + a^2 y '' + a_1 y ' + a_0 y = g(t).

(y[n], for example, represents the nth derivative of y; a_(n-1) is understood as a with subscript n - 1).

What is the higher-order equation?

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Your solution:

I really have no idea how to go about solving this problem???

y’_1 = y_2

y’_2 = y_3

y’_3 = -2y_1 + 4y_3 + e^(3t)

Not sure how to find y_1 and y_3???

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You've almost got it.

If you let y = y_1, then

y_2 = y_1 ' = y '

and

y_3 = y_2 ' = y ''.

So your equation

y_3 ' = -2 y_1 + 4 y_3 + e^(3t)

becomes

y ''' = -2 y + 4 y '' + e^(3t)

or

y ''' - 4 y '' + 2 y = e^(3t).

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confidence rating #$&*:

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0

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