STD 12 - 9. differential equations — JEE कक्षा 11 साइन्स — Question

$x + y = v$

==> $1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$ प्रतिस्थापित करने पर,

समीकरण $(i)$ से, $\cos v\,\left( {\frac{{dv}}{{dx}} - 1} \right) = 1$

==> $\cos v\,\frac{{dv}}{{dx}} = 1 + \cos v$ ==> $\frac{{\cos v}}{{1 + \cos v}}dv = dx$

==>$\left[ {\frac{{2{{\cos }^2}(v/2) - 1}}{{2{{\cos }^2}(v/2)}}} \right]\,dv = dx$ ==> $\left[ {1 - \frac{1}{2}{{\sec }^2}(v/2)} \right]\,dv = dx$

समाकलन करने पर, $v - \tan (v/2) = x + c$

$x + y - \tan \left( {\frac{{x + y}}{2}} \right) = x + c$ ==> $y = \tan \left( {\frac{{x + y}}{2}} \right) + c$.