Solve xdx , + ydy , = , xdy , - ,ydx x^2 , + , y^2 - Vidyadip

MCQ

Solve $xdx\, + ydy\, = \,\frac{{xdy\, - \,ydx}}{{{x^2}\, + \,{y^2}}}$

✓
$\frac{1}{2}\left( {{x^2} + {y^2}} \right) = {\tan ^{ - 1}}\left( {y/x} \right) + c$
B
$\frac{1}{2}\left( {{x^2} + {y^2}} \right) + {\tan ^{ - 1}}\left( {y/x} \right) + c = 0$
C
$\frac{1}{2}\left( {{x^2} - {y^2}} \right) = {\tan ^{ - 1}}\left( {y/x} \right) + c$
D
$\left( {{x^2} + {y^2}} \right) = {\tan ^{ - 1}}\left( {y/x} \right) + c$

✓

Answer

Correct option: A.

$\frac{1}{2}\left( {{x^2} + {y^2}} \right) = {\tan ^{ - 1}}\left( {y/x} \right) + c$

a
The $D.E.$ can be written as

$\frac{1}{2} \mathrm{d}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{d}\left\{\tan ^{-1}(\mathrm{y} / \mathrm{x})\right\}$

Integrating, we get

$\frac{1}{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\tan ^{-1}(\mathrm{y} / \mathrm{x})+\mathrm{c}$

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