MCQ
$\cos (x + y)\,dy = \,\,dx$ નો ઉકેલ મેળવો.
- ✓$y = \tan \,\left( {\frac{{x + y}}{2}} \right) + c$
- B$y + {\cos ^{ - 1}}\left( {\frac{y}{x}} \right) = c$
- C$y = x\,\,\sec \left( {\frac{y}{x}} \right) + c$
- Dએકપણ નહી.
Put $x + y = v$. Differentiate $1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$
Put these values in $(i),$ $\cos v\,\left( {\frac{{dv}}{{dx}} - 1} \right) = 1$
==> $\cos v\,\frac{{dv}}{{dx}} = 1 + \cos v$ ==> $\frac{{\cos v}}{{1 + \cos v}}dv = dx$
==>$\left[ {\frac{{2{{\cos }^2}(v/2) - 1}}{{2{{\cos }^2}(v/2)}}} \right]\,dv = dx$==>$\left[ {1 - \frac{1}{2}{{\sec }^2}(v/2)} \right]\,dv = dx$
Integrate, $v - \tan (v/2) = x + c$
$x + y - \tan \left( {\frac{{x + y}}{2}} \right) = x + c$ ==> $y = \tan \left( {\frac{{x + y}}{2}} \right) + c$.
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