Solution of $(D²+16)y=\cos 4x$ is
- A. $y=A\cos 4x+B\sin 4x+\frac{1}{8}\cos 4x$
- B. $y=A\cos 4x+B\sin 4x+\frac{x}{8}\sin 4x$
- C. $y=A\cos 4x+B\sin 4x+\frac{1}{8}\sin 4x$
- D. $y=A\cos 4x+B\sin 4x+\frac{x}{8}\cos 4x$
Answer: B. $y=A\cos 4x+B\sin 4x+\frac{x}{8}\sin 4x$
Explanation: Auxiliary equation is $D²+16=0 \Rightarrow D=\pm 4j$. The Particular Integral (PI) = $\frac{1}{D²+16}\cos 4x = \frac{x}{2D}\cos 4x = \frac{x}{2}\int\cos 4x dx = \frac{x\sin 4x}{8}$. Therefore, the General solution is $y = A\cos 4x+B\sin 4x+\frac{x}{8}\sin 4x$.