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course Mth 279
7/22
Query 20 Differential Equations*********************************************
Question: Using variation of parameters, solve the equation
y '' + y = sec(t), -pi/2 < t < pi/2.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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)
y(t) = C_1 cos(t) + C_2 sin(t) - cos(t) ln|cos(t)| + tsin(t)
if you take integral with respect to s how is it all of y_p is in respect to t??? wouldn’t
the solution techniquely be
y_p = -cos(t) ln|cos(s)| + s*sin(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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confidence rating #$&*:
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1
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Using variation of parameters, solve the equation
y '' + 36 y = csc^3 ( 6 t ).
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Your solution:
y_c?
r^2 + 36 = 0
r = +- 6i
y_c = C_1cos(6t) + C_2sin(6t)
y_p?
y_p = y_1u_1 + y_2u_2
y_p = cos(6t) u_1 + sin(6t)u_2
y’p = -6sin(6t)u’ + 6cos(6t)u’_2
W = | cos(6t) , sin(6t); -6sin(6t), 6cos(6t)| = 6cos^2(6t) - (-6sin^2(6t)) = 6(cos^2(6t) + sin^2(6t)) = 6
y_p = -cos(6t) int((sin(6t)csc^3(6t))/6) + sin(6t) int((cos(6t)csc^3(6t))/6)
int((sin(6t)csc^3(6t))/6) = int(csc^2(6t)/6) u = 6t du - 6
1/36 int(csc^2(u)) = 1/36 cot(6t)
int((cos(6t)csc^3(6t))/6) = 1/6 int(cot(6t)csc^2(6t)) u=6t du = 6
1/36int(cot(t)csc^2(u)) du w = cot(u) dw = -csc^2
-1/36 int(w dw)
-1/36 * w^2/2 = -cot^2(6t)/72
y_p = 1/36 * cos(6t)cot(6t) - 1/72 * sin(6t)cot^2(6t)
y(t) = C_1cos(6t) + C_2sin(6t) + 1/36 * cos(6t)cot(6t) - 1/72 * sin(6t)cot^2(6t)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
In general when and how would you find C_1 and C_2 in any of these types of problems from this assignment or that assignment before this???
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c_1 and c_2 are arbitrary constants. The solution satisfies the equation itself no matter what the values of those constants.
However if there are initial conditions, the constants are then determined, and are found in the usual manner.
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07-23-2014
&#Good responses. Let me know if you have questions. &#