MCQ
The general solution of the differential equation $\frac{{dy}}{{dx}} + \sin \left( {\frac{{x + y}}{2}} \right) = \sin \left( {\frac{{x - y}}{2}} \right)$ is
  • A
    $\log \tan \left( {\frac{y}{2}} \right) = c - 2\sin x$
  • $\log \tan \,\left( {\frac{y}{4}} \right) = c - 2\sin \left( {\frac{x}{2}} \right)$
  • C
    $\log \tan \,\left( {\frac{y}{2} + \frac{\pi }{4}} \right) = c - 2\sin x$
  • D
    $\log \tan \left( {\frac{y}{4} + \frac{\pi }{4}} \right) = c - 2\sin \left( {\frac{x}{2}} \right)$

Answer

Correct option: B.
$\log \tan \,\left( {\frac{y}{4}} \right) = c - 2\sin \left( {\frac{x}{2}} \right)$
b
(b) $\frac{{dy}}{{dx}} + \sin \left( {\frac{{x + y}}{2}} \right) = \sin \left( {\frac{{x - y}}{2}} \right)$

==> $\frac{{dy}}{{dx}} = \sin \left( {\frac{{x - y}}{2}} \right) - \sin \left( {\frac{{x + y}}{2}} \right)$

==> $\frac{{dy}}{{dx}} = - 2\sin \,\left( {\frac{y}{2}} \right)\,.\cos \,\left( {\frac{x}{2}} \right)$

==> ${\rm{cos}}{\rm{ec}}\left( {\frac{y}{2}} \right).dy = - 2\cos \left( {\frac{x}{2}} \right)\,dx$

Integrating both sides,

$\int {{\rm{cosec}}\left( {\frac{y}{{\rm{2}}}} \right)dy = - \int {2\cos \left( {\frac{x}{2}} \right)dx + c} } $.

==> $\frac{{\log \,\tan \frac{y}{4}}}{{1/2}} = - \frac{{2\sin \left( {x/2} \right)}}{{1/2}} + c$

==> $\log (\tan \frac{y}{4}) = c - 2\sin (x/2)$.

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