Question

The degree of tbe differential equation whose primitive is $c^2+2cy+a^2-x^2=0$, where c is an arbitrary and a is definite constant is:

Answer 1

$c^2+2cy+a^2-x^2=0$

Differentiating this

$2c\frac{dy}{dx}-2x=0\Rightarrow c=\frac{x}{\frac{dy}{dx}}$

Putting this value in the given equation, we have

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

Hence degree is $2$