dy dx = x + y x - y નો ઉકેલ મેળવો. - Vidyadip

MCQ

$\frac{{dy}}{{dx}} = \frac{{x + y}}{{x - y}}$ નો ઉકેલ મેળવો.

A
$c{({x^2} + {y^2})^{1/2}} + {e^{{{\tan }^{ - 1}}(y/x)}} = 0$
✓
$c{({x^2} + {y^2})^{1/2}} = {e^{{{\tan }^{ - 1}}(y/x)}}$
C
$c({x^2} - {y^2}) = {e^{{{\tan }^{ - 1}}(y/x)}}$
D
એકપણ નહી.

✓

Answer

Correct option: B.

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

(b) Given equation, $\frac{{dy}}{{dx}} = \frac{{x + y}}{{x - y}}$

It is a homogeneous equation so putting $y = vx$

and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}},$ we get

$v + x\frac{{dv}}{{dx}} = \frac{{x + vx}}{{x - vx}} = \frac{{1 + v}}{{1 - v}}$

==> $x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{1 - v}}$

==> $\frac{1}{x}dx = \left( {\frac{1}{{1 + {v^2}}} - \frac{v}{{1 + {v^2}}}} \right)dv$

==> ${\log _e}x = {\tan ^{ - 1}}v - \frac{1}{2}\log (1 + {v^2}) + {\log _e}c$

Substituting $v = \frac{y}{x},$we get

${\log _e}x = {\tan ^{ - 1}}\frac{y}{x} - \frac{1}{2}\log \left[ {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right] + {\log _e}c$

==> $c{({x^2} + {y^2})^{1/2}} = {e^{{{\tan }^{ - 1}}(y/x)}}$.

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