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

==> $1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$ ==> $\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} - 1$

 ${v^2}\left( {\frac{{dv}}{{dx}} - 1} \right) = {a^2}$

==> $\frac{{dv}}{{dx}} = \frac{{{a^2}}}{{{v^2}}} + 1 = \frac{{{a^2} + {v^2}}}{{{v^2}}}$ ==> $\frac{{{v^2}}}{{{a^2} + {v^2}}}dv = dx$

==> $\left( {1 - \frac{{{a^2}}}{{{a^2} + {v^2}}}} \right)dv = dx$ ==> $v - a{\tan ^{ - 1}}\frac{v}{a} = x + c$

==> $y = a{\tan ^{ - 1}}\left( {\frac{{x + y}}{a}} \right)+ c$