Topic: Engineering Mathematics
Solution of $(D^{2}+4)y=\sin^{2}x$ is
- A. $y=A~\cos2x+B~\sin2x-\frac{1}{8}-\frac{x}{8}\sin~2x$
- B. $y=A~\cos2x+B~\sin~2x+\frac{1}{8}+\frac{x}{8}\sin~2x$
- C. $y=A~\cos2x+B~\sin2x-\frac{1}{8}+\frac{x}{8}\sin~2x$
- D. $y=A~\cos2x+B~\sin2x+\frac{1}{8}-\frac{x}{8}\sin~2x$
Answer: D. $y=A~\cos2x+B~\sin2x+\frac{1}{8}-\frac{x}{8}\sin~2x$
Explanation: The particular solution is given by $\frac{1}{D^{2}+4}[\frac{1-\cos 2x}{2}]$. This evaluates to $\frac{1}{2}[\frac{1}{D^{2}+4}\cdot e^{0x}-\frac{1}{D^{2}+4}\cos 2x] = \frac{1}{8}-\frac{x}{4}\frac{\sin 2x}{2} = \frac{1}{8}-\frac{x}{8}\sin 2x$. The general solution is the sum of the complementary function $A \cos 2x + B \sin 2x$ and the particular solution.