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If $y=A \cos n x+B \sin n x$, then $y_2+n^2 y$ is equal to
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Verified Answer
The correct answer is:
$0$
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
$\Rightarrow \quad \begin{aligned} & y_2=-A n^2 \cos n x-B n^2 \sin n x \\ & \Rightarrow \quad y_2=-n^2(A \cos n x+B \sin n x) \\ & y_2=-n^2 \cdot y \Rightarrow y_2+n^2 y=0\end{aligned}$
$y=A \cos n x+B \sin n x$
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
$\Rightarrow \quad \begin{aligned} & y_2=-A n^2 \cos n x-B n^2 \sin n x \\ & \Rightarrow \quad y_2=-n^2(A \cos n x+B \sin n x) \\ & y_2=-n^2 \cdot y \Rightarrow y_2+n^2 y=0\end{aligned}$
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