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Question
\[\text { Find A and B so that y = A } \sin3x + B \cos3x \text { satisfies the equation }\]
\[\frac{d^2 y}{d x^2} + 4\frac{d y}{d x} + 3y = 10 \cos3x \] ?
Solution
\[\text { We have }, \]
\[y = A \sin3x + B \cos3x\]
\[\frac{d y}{d x} = 3A \cos3x - 3B \sin3x\]
\[\frac{d^2 y}{d x^2} = - 9A \sin3x - 9B \cos3x\]
\[\text{ Therefore, } \]
\[\frac{d^2 y}{d x^2} + 4\frac{d y}{d x} + 3y = - 9A \sin3x - 9B \cos3x + 12A \cos3x - 12B \sin3x + 3A \sin3x + 3B \cos3x\]
\[ = \left( - 6A - 12B \right) \sin3x + \left( - 6B + 12A \right) \cos3x\]
\[\text { It is given that}, \]
\[\frac{d^2 y}{d x^2} + 4\frac{d y}{d x} + 3y = 10 \cos3x\]
\[\text {Comparing the coefficients of }\sin3x \text { and } \cos3x, \text {we get }\]
\[ - 6A - 12B = 0 \text { and }- 6B + 12A = 10\]
\[\text { or A } = \frac{2}{3}\text { and B } = - \frac{1}{3}\]
\[\text { Hence, A } = \frac{2}{3} \text { and B } = - \frac{1}{3} .\]
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