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JEE Mains · Maths · STD 12 - 9. differential equations

ધારો કે \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) એ  વિકલ સમીકરણ\(\sqrt{1-\mathrm{x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{1-\mathrm{y}^{2}}=0,|\mathrm{x}|<1\) નો ઉકેલ આપેલ છે . જો  \(\mathrm{y}\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2},\) હોય તો  \(\mathrm{y}\left(\frac{-1}{\sqrt{2}}\right)\) મેળવો.

  1. A \(-\frac{\sqrt{3}}{2}\)
  2. B \(\frac{1}{\sqrt{2}}\)
  3. C \(\frac{\sqrt{3}}{2}\)
  4. D \(-\frac{1}{\sqrt{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{\sqrt{2}}\)

Step-by-step Solution

Detailed explanation

\(\frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}\) so, \(\frac{d y}{\sqrt{1-y^{2}}}+\frac{d x}{\sqrt{1-x^{2}}}=0\) Integrating, \(\sin ^{-1} x+\sin ^{-1} y=c\) so, \(\frac{\pi}{6}+\frac{\pi}{3}=c\) Hence, \(\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}\) Put…
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