I got a particular solution y_p(x) that is different from what the book has.(adsbygoogle = window.adsbygoogle || []).push({});

y'' + 9y = 2cos3x + 3sin3x

Characteristic equation: r^2 + 9 = 0

(r+3i)(r-3i) = 0

y_c = c_1cos3x + c_2sin3x

y_p = Acos3x + Bsin3x (not linearly independent, so I'll try another y_p)

y_p = Axcos3x + Bxsin3x

y'_p = Acos3x - 3Axsin3x + Bsin3x + 3Bxcos3x

y''_p = -3Asin3x - 3Asin3x - 9Axcos3x + 3Bcos3x + 3Bcos3x - 9Bxsin3x

x = 0

y(x) = Acos3x + Bsin3x - 6Asin3x + 6Bcos3x + Acos3x + Bsin3x - 6Asin3x + 6Bcos3x = 2cos3x + 3sin3x

Acos3x + 6Bcos3x = 2cos3x

A + 6B = 1

Bsin3x - 6Asin3x = 3sin3x

-12A + 2B = 3

My ti89 says A = -8/37 , B = 15/74

y_p = (15/37sin3x - 8/37xcos3x)

But the book's answer is y_p = 1/6(2xsin3x - 3xcos3x)

What have I missed?

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# Homework Help: Nonhomogeneous Differential Equation

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