Q71 / 155 · WBJEE 2017 · Mathematics

If $y=e^{m \sin ^{-1} x}$ then $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0,$ where $k$ is equal to

  • Correct answer
    A
    $m^{2}$
  • B
    2
  • C
    -1
  • D
    $-m^{2}$

Solution

$y=e^{msin^{-1}x}$ $\Rightarrow \quad \frac{d y}{d x}=e^{msin^{-1}x} \cdot \frac{m}{\sqrt{1-x^{2}}}$ $\Rightarrow \sqrt{1-x^{2}} \frac{d y}{d x}=m e^{msin^{-1}x}$ $\Rightarrow \sqrt{1-x^{2}} \frac{d y}{d x}=m y$ $\Rightarrow \sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}}+\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \frac{d y}{d x}=m \frac{d y}{d x}$ $\Rightarrow\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=m \sqrt{1-x^{2}} \frac{d y}{d x}$ $\Rightarrow\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=m \cdot(m y)$ $\Rightarrow\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-m^{2} y=0$ $\therefore$ $k=m^{2}$

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

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