Query 23

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course MTH 279

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:

y_1’ = (3t)/(t+2) y_1 + 5/(t+2) y_2

y_2’ = 2/(t-2) y_1 + 4/(t-2) y_2

The continuity of the system is dependent on the individual continuities.

We find that both (3t)/(t+2) and 5/(t+2) are continuous on the intervals (-∞,-2)ᴗ(-2,∞).

We find that both 2/(t-2) and 4/(t-2) are continuous on the intervals (-∞,2)ᴗ(2,∞).

Thus, our system is continous on (-∞,-2)ᴗ(-2,2)ᴗ(2,∞).

Our t_0 = 1, which falls on the interval (-2,2), so our largest interval of definition is (-2,2).

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: 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:

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

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

Anti-differentiating:

y(t) = [y_1;y_2;y_3] = [y; y’; y”]

y’’’ = -2y_1 + 4y_3 + e^(3t)

y’’’ - 4y_3 + 2y_1 = e^(3t)

y_3 = y”

y_1 = y

y’’’ - 4y” + 2y = e^(3t)