9. differential equations — ગણિત ધોરણ 12 સાયન્સ — Question

==> $y(1 - ay) = \left( {a + x} \right)\,.\frac{{dy}}{{dx}}$ ==> $\frac{{dx}}{{(a + x)}} = \frac{{dy}}{{y(1 - ay)}}$

Integrating both sides, $\int {\frac{{dx}}{{(a + x)}} = } \int {\frac{{dy}}{{y(1 - ay)}}} $

==> $\int {\frac{{dx}}{{a + x}} = \int {\left[ {\frac{1}{y} + \frac{a}{{(1 - ay)}}} \right]\,dx} } $

$\log (a + x) = \log y + \frac{{a\log (1 - ay)}}{{ - a}}$

==> $\log (a + x) = \log y - \log (1 - ay) + \log c$

==> $\log (x + a)(1 - ay) = \log cy$ ==> $(x + a)(1 - ay) = cy$.