**Toolbox:**

- If the given equation has n arbitary constans, then its differential equation will be of $n^{th}$ order
- $\large\frac{d}{dx}$$(uv)=\large\frac{du}{dx}$$.v+\large\frac{dv}{dx}$$.u$

Given $Ax^2+By^2=1$-----(1)

We are asked to find the differential equation by eliminating A and B

Consider $Ax^2+By^2=1$

Differentiate with respect to x

$2Ax+2By.y'=0$=>$Ax=-Byy'$-----(2)

Again differentiate with respect to x

$yy'$ is in the product form, hence apply product rule

$\large\frac{d}{dx}$$(uv)=\large\frac{d}{dv}$$(u).v+\large\frac{d}{dx}$$(v).u$

Here $u=y$ hence $u'=y'$

and $v=y'$ hence $v'=y''$

Therefore $\large\frac{d}{dx}$$(yy')=y.y''+y'.y'$

$=(y')^2+y.y''$

Therefore differentiating $Ax=-Byy'$ we get

$A=-B[(y')^2+y.y'']$-----(3)

Now substituting the value of A in equ(2)

$-B[(y')^2+y.y'']x=-Byy''$

=>$x(y')^2+y.y''x=yy''$

=> $x(y')^2+y.y''x-yy''=0$

or => $x(y')^2+y.y''(x-1)=0$