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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
$y = ke^{\sin ^{-1} x} + 3$ is a solution of differeulial equation :-
  • ✓
    $\sqrt {1 - {x^2}} \frac{{dy}}{{dx}} = y - 3$
  • B
    $\sqrt {1 + {x^2}} \frac{{dy}}{{dx}} = y - 3$
  • C
    $\sqrt {1 + {x^2}} \frac{{dy}}{{dx}} = y + 3$
  • D
    $\sqrt {1 - {x^2}} \frac{{dy}}{{dx}} = y + 3$
✓

Answer

Correct option: A.
$\sqrt {1 - {x^2}} \frac{{dy}}{{dx}} = y - 3$
a
$\frac{d y}{d x}=k e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}}$

$\frac{d y}{d x}=(y-3) \frac{1}{\sqrt{1-x^{2}}}$

$\sqrt{1-x^{2}} \frac{d y}{d x}=y-3$

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