Assignment 22

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course Mth 279

7/28

Query 20 Differential Equations*********************************************

Question: Using variation of parameters, solve the equation

y '' + y = sec(t), -pi/2 < t < pi/2.

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Your solution:

y_c?

r^2 + 1 = 0

r = +- 1i

y_c = C_1 cos(t) + C_2 sin(t)

y_p?

g(t) = sec(t)

can’t use the table b/c g(t) isn’t on there

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You aren't restricted to the table.

However there's no simple function the combination of whose derivatives can give you the secant. Some trial and error should convince you of this.

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general solution is

y_p = -y_1(t) int(y_2*g/w) + y_2 int(y_1*g/w)

y_p = y_1 u_1 + y_2 u_2

y_p = cos(t) u_1 + sin(t)u_2

y’_p = -sin(t)u’_1 + cos(t)u’_2 = 0

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The derivative of your y_p function is

y_p ' = -sin(t) u_1 ' - cos(t) u_1 + cos(t) u_2' - sin(t) u_2.

However we get to make the assumption that u_1 ' y_1 + u_2 ' y_2 = 0, which then gives us

y_p' = cos(t) u_1 + sin(t)u_2.

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From this we conclude that

y_p = -y_1(t) int(y_2*g/w) + y_2 int(y_1*g/w)

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W(t) = |cos(t), sin(t); -sin(t); cos(t) | = cos^2(t) - -sin^2(t) = cos^2 + sin^2 = 1

y_p = -cos(t) int((sin(s) * sec(s))/1)ds + sin(t) int((cos(s)*sec(s))/1)ds

= -cos(t) int(sin/cos) + sin(t) int(1)

let u = cos du= sin

y_p = -cos(t) int(1/u)dt

y_p = -cos(t) ln|cos(t)| + tsin(t)

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There is no variable s in the solution.

The integrals are with respect to t, and you get

y_P = -ln | cos(t) | * cos(t) + t sin(t).

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y(t) = C_1 cos(t) + C_2 sin(t) - cos(t) ln|cos(t)| + tsin(t)

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: Using variation of parameters, solve the equation

y '' + 36 y = csc^3 ( 6 t ).

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Your solution:

y_c?

r^2 + 36 = 0

r = +- 6i

y_c = C_1cos(6t) + C_2sin(6t)

y_p?

y_p = y_1u_1 + y_2u_2

y_p = cos(6t) u_1 + sin(6t)u_2

y’p = -6sin(6t)u’ + 6cos(6t)u’_2

W = | cos(6t) , sin(6t); -6sin(6t), 6cos(6t)| = 6cos^2(6t) - (-6sin^2(6t)) = 6(cos^2(6t) + sin^2(6t)) = 6

y_p = -cos(6t) int((sin(6t)csc^3(6t))/6) + sin(6t) int((cos(6t)csc^3(6t))/6)

int((sin(6t)csc^3(6t))/6) = int(csc^2(6t)/6) u = 6t du - 6

1/36 int(csc^2(u)) = 1/36 cot(6t)

int((cos(6t)csc^3(6t))/6) = 1/6 int(cot(6t)csc^2(6t)) u=6t du = 6

1/36int(cot(t)csc^2(u)) du w = cot(u) dw = -csc^2

-1/36 int(w dw)

-1/36 * w^2/2 = -cot^2(6t)/72

y_p = 1/36 * cos(6t)cot(6t) - 1/72 * sin(6t)cot^2(6t)

y(t) = C_1cos(6t) + C_2sin(6t) + 1/36 * cos(6t)cot(6t) - 1/72 * sin(6t)cot^2(6t)

confidence rating #$&*:

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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:

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