Differential Equations — MATHS STD 12 Science — Question

To prove:y given by eq. (i) is a solution of differential equation $y' = \frac{{{y^2}}}{{1 - xy}}$ ....(ii)

Proof: Differentiating both sides of eq. (i) w.r.t x, we have

$xy' + y\left( 1 \right) = \frac{1}{y}y' + 0$

$ \Rightarrow xy' - \frac{{y'}}{y} = - y$

$ \Rightarrow y'\left( {x - \frac{1}{y}} \right) = - y$

$ \Rightarrow y'\left( {\frac{{xy - 1}}{y}} \right) = - y$

$\Rightarrow y'\left( {xy - 1} \right) = - {y^2}$

$ \Rightarrow y' = \frac{{ - {y^2}}}{{xy - 1}}$

$ \Rightarrow y' = \frac{{ - {y^2}}}{{ - \left( {1 - xy} \right)}} = \frac{{{y^2}}}{{1 - xy}}$

Hence, function (implicit) given by eq. (i) is a solution of $y' = \frac{{{y^2}}}{{1 - xy}}$.