Differential Equations — Class 12 Maths Solution

Given differential equation is $A{x^2} + B{y^2} = 1$
On differentiating both sides w.r.t. $x$,

we get
$2Ax + 2By\frac{{dy}}{{dx}} = 0$

$\Rightarrow$2 By $\frac{{dy}}{{dx}} = - 2Ax$

$\Rightarrow$ By $\frac{{dy}}{{dx}} = - Ax \Rightarrow \frac{y}{x} \cdot \frac{{dy}}{{dx}} = - \frac{A}{B}$

Again, differentiating w.r.t. $x$,

we get
$\frac{y}{x} \cdot \frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} \cdot \left( {\frac{{x\frac{{dy}}{{dx}} - y}}{{{x^2}}}} \right) = 0$

$\Rightarrow$ $\frac{y}{x} \cdot \frac{{{d^2}y}}{{d{x^2}}} + \frac{{x{{\left( {\frac{{dy}}{{dx}}} \right)}^2} - y\left( {\frac{{dy}}{{dx}}} \right)}}{{{x^2}}} = 0$

$\Rightarrow$ $xy\frac{{{d^2}y}}{{d{x^2}}} + x{\left( {\frac{{dy}}{{dx}}} \right)^2} - y\left( {\frac{{dy}}{{dx}}} \right) = 0$

$\Rightarrow$ $xy\,{y^{\prime \prime }} + x{\left( {{y^\prime }} \right)^2} - y{y^\prime } = 0$