Topic: Engineering Mathematics
Solution of $(D^{2}+4)y=sin^{2}x,$ is
- A. $y=A~cos~2x+B~sin~2x-\frac{1}{8}-\frac{x}{8}sin~2x$
- B. $y=A~cos~2x+B~sin~2x+\frac{1}{8}+\frac{x}{8}sin~2x$
- C. $y=A~cos~2x+B~sin~2x-\frac{1}{8}+\frac{x}{8}sin~2x$
- D. $y=A~cos~2x+B~sin~2x+\frac{1}{8}-\frac{x}{8}sin~2x$
Answer: D. $y=A~cos~2x+B~sin~2x+\frac{1}{8}-\frac{x}{8}sin~2x$
Explanation: Auxiliary equation is $D^{2}+4=0 \implies D=\pm2i$. So Complementary Function (CF) is $A~cos~2x+B~sin~2x$. Particular Integral (PI) $= \frac{1}{D^{2}+4}sin^{2}x = \frac{1}{D^{2}+4}(\frac{1-cos~2x}{2}) = \frac{1}{2(D^{2}+4)}(e^{0x}) - \frac{1}{2(D^{2}+4)}cos~2x = \frac{1}{2(4)} - \frac{x}{2(2\cdot 2)}sin~2x = \frac{1}{8} - \frac{x}{8}sin~2x$. Complete solution is $y = CF + PI = A~cos~2x+B~sin~2x+\frac{1}{8}-\frac{x}{8}sin~2x$.