The solution of the differential equation (1 + x^2 ) dy dx = x is - Vidyadip

MCQ

The solution of the differential equation $(1 + {x^2})\frac{{dy}}{{dx}} = x$ is

A
$y = {\tan ^{ - 1}}x + c$
B
$y = - {\tan ^{ - 1}}x + c$
✓
$y = \frac{1}{2}{\log _e}(1 + {x^2}) + c$
D
$y = - \frac{1}{2}{\log _e}(1 + {x^2}) + c$

✓

Answer

Correct option: C.

$y = \frac{1}{2}{\log _e}(1 + {x^2}) + c$

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

==> $\int {dy} = \int {\frac{x}{{1 + {x^2}}}dx} + c$ ==> $y = \frac{1}{2}{\log _e}(1 + {x^2}) + c$.

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