Q118 / 200 · AP EAMCET 2001 · Mathematics

If $y=A \cos n x+B \sin n x$, then $y_2+n^2 y$ is equal to

  • Correct answer
    A
    $0$
  • B
    $1$
  • C
    $y$
  • D
    $-1$

Solution

We have, $y=A \cos n x+B \sin n x$ On differentiating w.r.t. $x$, we get $y_1=-A n \sin n x+B n \cos n x$ Again differentiating w.r.t. $x$, we get $\Rightarrow \quad \begin{aligned} & y_2=-A n^2 \cos n x-B n^2 \sin n x \\ & \Rightarrow \quad y_2=-n^2(A \cos n x+B \sin n x) \\ & y_2=-n^2 \cdot y \Rightarrow y_2+n^2 y=0\end{aligned}$

Hence, the correct answer is option: 'A'.

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