shiri
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Consider the equation y''+4y=2sin(2x)+e^{2x}+2. According to the method of Undetermined Coefficients, the particular solution has the form:
C1 = 1st constant
C2 = 2nd constant
C3 = 3rd constant
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.
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A) C1xsin(2x) + C3x + C4xe^{2x} + C5xe^{-2x}
B) C1sin(2x) + (C2)2 + C3xe^{-2x}
C) C1xcos(2x) + C2xsin(2x) + C3 + C4e^{2x}
D) C1cos(2x) + C2xsin(2x) + C3 + C4xe^{2x}
Solution:
y^2 + 16 = 0
y = -4, 4
yh = c1cos(x) + c2sin(x)
yp = Acos(2x) + Bsin(2x) + C + De^{2x}
y'p = -2Asin(2x) + 2Bcos(2x) + 2De^{2x}
y''p = -4Acos(2x) - 4Bsin(2x) + 4De^{2x}
y'' + 4y = 2sin(2x) + 2 e^{2x}
[-4Acos(2x) - 4Bsin(2x) + 4De^{2x}] + 16[Acos(2x) + Bsin(2x) + C + De^{2x}] = 2sin(2x) + 2 + e^{2x}
12Acos(2x) + 12Bsin(2x) + 20De^{2x} + 16C = 2sin(2x) + 2 + e^{2x}
cos(2x) ===> 12A = 0 ===> A = 0
sin(2x) ===> 12B = 2 ===> B = 1/6
16 ===> 16C = 2===> C = 1/8
e^{2x} ===> 20D = 1 ===> D = 1/20
So far what I got B) as a right answer, but I'm not too sure about A), C), and D)
Can anybody help me out here, please?
C1 = 1st constant
C2 = 2nd constant
C3 = 3rd constant
.
.
.
A) C1xsin(2x) + C3x + C4xe^{2x} + C5xe^{-2x}
B) C1sin(2x) + (C2)2 + C3xe^{-2x}
C) C1xcos(2x) + C2xsin(2x) + C3 + C4e^{2x}
D) C1cos(2x) + C2xsin(2x) + C3 + C4xe^{2x}
Solution:
y^2 + 16 = 0
y = -4, 4
yh = c1cos(x) + c2sin(x)
yp = Acos(2x) + Bsin(2x) + C + De^{2x}
y'p = -2Asin(2x) + 2Bcos(2x) + 2De^{2x}
y''p = -4Acos(2x) - 4Bsin(2x) + 4De^{2x}
y'' + 4y = 2sin(2x) + 2 e^{2x}
[-4Acos(2x) - 4Bsin(2x) + 4De^{2x}] + 16[Acos(2x) + Bsin(2x) + C + De^{2x}] = 2sin(2x) + 2 + e^{2x}
12Acos(2x) + 12Bsin(2x) + 20De^{2x} + 16C = 2sin(2x) + 2 + e^{2x}
cos(2x) ===> 12A = 0 ===> A = 0
sin(2x) ===> 12B = 2 ===> B = 1/6
16 ===> 16C = 2===> C = 1/8
e^{2x} ===> 20D = 1 ===> D = 1/20
So far what I got B) as a right answer, but I'm not too sure about A), C), and D)
Can anybody help me out here, please?
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