y^2 ,dx + ( x^2 - xy + y^2 ) , ,dy = 0 નો ઉકેલ મેળવો. - Vidyadip

MCQ

${y^2}\,dx + ({x^2} - xy + {y^2})\,\,dy = 0$ નો ઉકેલ મેળવો.

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

✓

Answer

Correct option: A.

${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) + \log y + c = 0$

a
(a) $\frac{{dx}}{{dy}} + \frac{{{x^2} - xy + {y^2}}}{{{y^2}}} = 0$

$\frac{{dx}}{{dy}} + {\left( {\frac{x}{y}} \right)^2} - \left( {\frac{x}{y}} \right) + 1 = 0$

Put $v = x/y$ ==> $x = vy$ ==> $\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}$

$v + y\frac{{dv}}{{dy}} + {v^2} - v + 1 = 0$ ==> $\frac{{dv}}{{{v^2} + 1}} + \frac{{dy}}{y} = 0$

==> $\int {\frac{{dv}}{{{v^2} + 1}} + \int {\frac{{dy}}{y} = 0} } $ ==> ${\tan ^{ - 1}}(v) + \log y + C = 0$

==> ${\tan ^{ - 1}}(x/y) + \log y + c = 0$.

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