Question

The degree of the differential equation satisfying the equation $\frac{x^2}{a^2+\lambda }+\frac{y^2}{b^2+\lambda }=1$ where $a$ and $b$ are specified constants and $\lambda$ is an arbitrary parameter, is

Answer 1

$\frac{x^2}{a^2+\lambda }+\frac{y^2}{b^2+\lambda }=1\cdots \left(i\right)$
Differenting w.r.t $x,$
$\Rightarrow \frac{2x}{a^2+\lambda }+\frac{2y}{b^2+\lambda }\cdot \frac{dy}{dx}=0\frac{x}{a^2+\lambda }=-\frac{y}{b^2+\lambda }\cdot \frac{dy}{dx}\cdots \left(ii\right)$
Substituting in $\left(i\right),$
$\Rightarrow -\frac{xy}{b^2+\lambda }\cdot \frac{dy}{dx}+\frac{y^2}{b^2+\lambda }=1\Rightarrow b^2+\lambda =y^2-xy\frac{dy}{dx}\Rightarrow \lambda =y^2-xy\frac{dy}{dx}-b^2$
from $\left(i\right),$
$\frac{x^2}{a^2-b^2+y^2-xy\frac{dy}{dx}}+\frac{y^2}{y^2-xy\frac{dy}{dx}}=1\Rightarrow x^2(y^2-xy\frac{dy}{dx})+y^2(a^2-b^2+y^2-xy\frac{dy}{dx})=(a^2-b^2+y^2-xy\frac{dy}{dx})(y^2-xy\frac{dy}{dx})$
in the above equation,
Degree of $\frac{dy}{dx}$ is $2.$
$\therefore$ degree of $D.E.$ is $2.$