Topic: Engineering Mathematics
Solution of $(D^{2}+16)y=cos~4x$ is-
- A. $y=Acos~4x+Bsin~4x+\frac{1}{8}cos~4x$
- B. $y=Acos~4x+B~sin~4x+\frac{1}{8}sin~4x$
- C. $y=Acos~4x+Bsin~4x+\frac{x}{8}cos~4x$
- D. $y=Acos~4x+B~sin~4x+\frac{x}{8}sin~4x$
Answer: D. $y=Acos~4x+B~sin~4x+\frac{x}{8}sin~4x$
Explanation: The Complementary Function (C.F.) is found by $(D^{2}+16) = 0 \Rightarrow D = \pm 4i$. So, $C.F. = A \cos 4x + B \sin 4x$. The Particular Integral (P.I.) is $\frac{1}{D^{2}+16}\cos 4x$. Since $-a^{2} = -4^{2} = -16$, substituting directly gives zero in the denominator. Using the rule $\frac{1}{D^{2}+a^{2}}\cos ax = \frac{x}{2a}\sin ax$, here $a=4$, so $P.I. = \frac{x}{8}\sin 4x$. Total solution $y = C.F. + P.I. = A \cos 4x + B \sin 4x + \frac{x}{8}\sin 4x$.