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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
If $\frac{{dy}}{{dx}} + \frac{1}{{\sqrt {1 - {x^2}} }} = 0$, then
  • ✓
    $y + {\sin ^{ - 1}}x = c$
  • B
    ${y^2} + 2{\sin ^{ - 1}}x + c = 0$
  • C
    $x + {\sin ^{ - 1}}y = 0$
  • D
    ${x^2} + 2{\sin ^{ - 1}}y = 1$
✓

Answer

Correct option: A.
$y + {\sin ^{ - 1}}x = c$
a
(a) $\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^2}} }}$==> $dy = - \frac{1}{{\sqrt {1 - {x^2}} }}dx$

On integrating, we get $y = {\cos ^{ - 1}}x + c$

==> $y = \frac{\pi }{2} - {\sin ^{ - 1}}x + c$ ==> $y + {\sin ^{ - 1}}x = c$.

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