STD 12 - 9. differential equations — JEE STD 11 Science — Question

Put $x + y = t$ ==> $\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} - 1$

$\therefore \frac{{dy}}{{dx}} = \frac{{1 - t}}{{2t - 3}}$  ==> $\frac{{dt}}{{dx}} - 1 = \frac{{1 - t}}{{2t - 3}}$ ==> $\frac{{dt}}{{dx}} = \frac{{t - 2}}{{2t - 3}}$

==> $\frac{{2t - 3}}{{t - 2}}dt = dx$. Integrating both sides, we get
$\int_{}^{} {\frac{{2t - 4}}{{t - 2}}dt} - \int_{}^{} {\frac{{3 - 4}}{{t - 2}}dt} = \int_{}^{} 1 dx$

==> $2t + \log (t - 2) = x + c$

==> $2(x + y) + \log (x + y - 2) = x + c$

==> $2y + x + \log (x + y - 2) = c$.