Question:
The particular integral of \( \frac{d^2y}{dx^2} + y = \cos(2x - 1) \) is
The particular integral of \( \frac{d^2y}{dx^2} + y = \cos(2x - 1) \) is
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Use operator method with \( f(D) \) when solving linear differential equations with trigonometric RHS.
Updated On: May 29, 2025
- \( \frac{1}{65} (\cos(2x - 8\sin 2x)) \)
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- \( \frac{1}{65} (\cos(2x + 8\sin 2x)) \)
- \( \frac{1}{65} (\cos(2x - 1) - 8\sin(2x - 1)) \)
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The Correct Option is D
Solution and Explanation
Let \( f(D)y = \cos(2x - 1) \), where \( f(D) = D^2 + 1 \)
Then PI = \( \frac{1}{f(D)} \cos(2x - 1) \)
Using: \( \frac{1}{D^2 + 1} \cos(ax + b) = \frac{D^2 + 1}{(D^2 + 1)^2 + 4a^2} \cos(ax + b) \),
Simplifies to: \( \frac{1}{(4 - (-1))^2 + 1} = \frac{1}{65} (\cos(2x - 1) - 8\sin(2x - 1)) \)
Then PI = \( \frac{1}{f(D)} \cos(2x - 1) \)
Using: \( \frac{1}{D^2 + 1} \cos(ax + b) = \frac{D^2 + 1}{(D^2 + 1)^2 + 4a^2} \cos(ax + b) \),
Simplifies to: \( \frac{1}{(4 - (-1))^2 + 1} = \frac{1}{65} (\cos(2x - 1) - 8\sin(2x - 1)) \)
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