Differential equation of y = ( ^ - 1 x) is - Vidyadip

MCQ

Differential equation of $y = \sec ({\tan ^{ - 1}}x)$ is

A
$(1 + {x^2})\frac{{dy}}{{dx}} = y + x$
B
$(1 + {x^2})\frac{{dy}}{{dx}} = y - x$
✓
$(1 + {x^2})\frac{{dy}}{{dx}} = xy$
D
$(1 + {x^2})\frac{{dy}}{{dx}} = \frac{x}{y}$

✓

Answer

Correct option: C.

$(1 + {x^2})\frac{{dy}}{{dx}} = xy$

c
(c) $y = \sec ({\tan ^{ - 1}}x)$

$\frac{{dy}}{{dx}} = \sec ({\tan ^{ - 1}}x)\tan ({\tan ^{ - 1}}x)\,.\,\frac{1}{{1 + {x^2}}}$$ = \frac{{xy}}{{1 + {x^2}}}$

==> $(1 + {x^2})\frac{{dy}}{{dx}} = xy$.

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