Query 19

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course MTH 279

4/2 7pm

Query 19 Differential Equations*********************************************

Question: Find the general solution of the equation

y '' + y = e^t sin(t).

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Your solution:

y = 1/5e^tsint - 2/5e^tcost + C_1cost + C_2sint

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 general solution of the equation

y '' + y ' = 6 t^2

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Your solution:

y = 2t^3 - 6t^2 +12t + C_1e^(-.5t) + C_2e^(.5t)

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating:

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Question: Find the general solution of the equation

y '' + y ' = cos(t).

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Your solution:

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: Give the expected form of the particular solution to the given equation, but do not actually solve for the constants:

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

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Your solution:

y = Ae^-tsint + Be^-tcost + Ct^2 + Dt + E + Ft^2e^3t

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The complementary solution is A e^(3 t) + B e^(-t)

The term 2 e^-t cos(t) is expected to result from a combination of multiples of e^-t cos(t) and e^(-t) sin (t).

The term t^2 is expected to result from a second-degree polynomial of the form a t^2 + b t + c, with a = 1. The constants b and c can be adjusted so that the various derivatives of the other terms cancel.

The term t e^(3 t) would be expected to result from a combination of multiples of t e^(3 t) and e^(3 t), except that e^(3 t) is a solution of the homogeneous equation. So we will try a combination of t^2 e^(3 t) and t e^(3 t).

*@

confidence rating #$&*:

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

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Give the expected form of the particular solution to the given equation, but do not actually solve for the constants:

y '' + 4 y = 2 sin(t) + cosh(t) + cosh^2(t).

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Your solution:

y = Acost + Bsint + .5Ce^t + .5De^-t + .25e^2t + .25e^-2t + .5

confidence rating #$&*:

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

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating:

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Question: The equation

y '' + alpha y ' + beta y = t + sin(t)

has complementary solution y_C = c_1 cos(t) + c_2 sin(t) (i.e., this is the solution to the homogeneous equation).

Find alpha and beta, and solve the equation.

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Your solution:

alpha = 0

beta = 1

y = t - .5tcost + C_1cost + C_2sint

confidence rating #$&*:

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

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Consider the equation

y '' - y = e^(`i * 2 t),

where `i = sqrt(-1).

Using trial solution

y_P = A e^(i * 2 t)

find the value of A, which is in general a complex number (though in some cases the real or imaginary part of A might be zero)

Show that the real and imaginary parts of the resulting function y_P are, respectively, solutions to the real and imaginary parts of the original equation.

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Your solution:

A = -1/5

y_P = -1/5e^(i2t) = -1/5(cost2t + isin2t)

original problem rewritten: y'' - ycos2t +isin2t

Real parts:

(-1/5cos2t)'' - (-1/5cos2t) = cos2t

yes

Imaginary parts:

(-1/5isin2t)'' - (-1/5isin2t) = isin2t

yes

confidence rating #$&*:

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

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: "

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