5. continuity and differentiation — ગણિત ધોરણ 12 સાયન્સ — Question

==> $2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2y\frac{{{d^2}y}}{{d{x^2}}} = 2a $

$\Rightarrow y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{dy}}{{dx}}} \right)^2}$

==>$y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{2ax + b}}{{2y}}} \right)^2}$

==> $y\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4a{y^2} - {{(2ax + b)}^2}}}{{4{y^2}}}$

==> $4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4a(a{x^2} + bx + c) - (4{a^2}{x^2} + 4abx + {b^2})$

==>$4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4ac - {b^2} \Rightarrow {y^3}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4ac - {b^2}}}{4} = $a constant.