The solution of the differential equation dy dx + 1 + x^2 x = 0 i… - Vidyadip

MCQ

The solution of the differential equation $\frac{{dy}}{{dx}} + \frac{{1 + {x^2}}}{x} = 0$ is

A
$y = - \frac{1}{2}{\tan ^{ - 1}}x + c$
✓
$y + \log x + \frac{{{x^2}}}{2} + c = 0$
C
$y = \frac{1}{2}{\tan ^{ - 1}}x + c$
D
$y - \log x - \frac{{{x^2}}}{2} = c$

✓

Answer

Correct option: B.

$y + \log x + \frac{{{x^2}}}{2} + c = 0$

b
(b) $\frac{{dy}}{{dx}} + \frac{{1 + {x^2}}}{x} = 0$ ==> $dy + \left( {\frac{1}{x} + x} \right){\rm{ }}dx = 0$

On integrating, we get $y + \log x + \frac{{{x^2}}}{2} + c = 0$.

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