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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
Solution of differential equation $\frac{{dy}}{{dx}} = \frac{{y - x}}{{y + x}}$ is
  • ✓
    ${\log _e}({x^2} + {y^2}) + 2{\tan ^{ - 1}}\frac{y}{x} + c = 0$
  • B
    $\frac{{{y^2}}}{2} + xy = xy - \frac{{{x^2}}}{2} + c$
  • C
    $\left( {1 + \frac{x}{y}} \right){\rm{ }}y = \left( {1 - \frac{x}{y}} \right){\rm{ }}x + c$
  • D
    $y = x - 2{\log _e}y + c$
✓

Answer

Correct option: A.
${\log _e}({x^2} + {y^2}) + 2{\tan ^{ - 1}}\frac{y}{x} + c = 0$
a
(a) Put $y = vx$ ==> $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

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

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

==> $ - {\log _e}x = \frac{1}{2}\int_{}^{} {\frac{{2v}}{{{v^2} + 1}}} dv + \int_{}^{} {\frac{1}{{{v^2} + 1}}} dv$

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

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

==> ${\log _e}({x^2} + {y^2}) + 2{\tan ^{ - 1}}\frac{y}{x} + c = 0$.

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