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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

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

If y=Acos⁡nx+Bsin⁡nxy=A \cos n x+B \sin n x, then y2+n2yy_2+n^2 y is equal to

Options

  1. A00
  2. B11
  3. Cyy
  4. D−1-1

Correct answer

A. 00

Step-by-step solution

We have, y=Acos⁡nx+Bsin⁡nxy=A \cos n x+B \sin n x On differentiating w.r.t. xx, we get y1=−Ansin⁡nx+Bncos⁡nxy_1=-A n \sin n x+B n \cos n x Again differentiating w.r.t. xx, we get ⇒\Rightarrow \quady2=−An2cos⁡nx−Bn2sin⁡nx⇒y2=−n2(Acos⁡nx+Bsin⁡nx)y2=−n2⋅y⇒y2+n2y=0\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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