Query_23

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course Mth 279

5/17

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Putting the above equations into standard form yeilds:

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

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

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

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

y(t) = [ y1(t); y2(t) ]

p(t) = [ 3t/(t+2), 5/(t+2); 2/(t-2), 4/(t-2)]

t_0 = 1

y_0 = [ 0; 2]

undefined for p11(t), and p12(t) on t = -2

Also for p21(t) and p22(t) on t = 2

Largest interval in which the solution is defined is either

(-infinity, -2) or (2, infinity)

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: 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:

Since Y(t) = [ Y1(t); Y2(t); Y3(t)] = [ y(t); y'(t); y''(t)]

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

rearranging we get:

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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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:

#*&!

&#Very good responses. Let me know if you have questions. &#