Differential equation degree doubt $$\frac{dy}{dx} = \sin^{-1} (y)$$ The above equation is a form of $\frac{dy}{dx} = f(y)$, so degree should be $1$.  But if I write it as $$y = \sin\left(\frac{dy}{dx}\right)$$ then degree is not defined as it is not a polynomial in $\frac{dy}{dx}$. Please explain?

To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example, 
$\frac{dy}{dx} = y, \frac{dy}{dx} = y^2 + 1$ are of the first order. 
$ \frac{d^2 y}{dx^2} = y, \frac{d^2 y}{dx^2} = (\frac{dy}{dx})^3 + 1$ are of second order.
Your differential equation $\frac{dy}{dx} = \sin^{-1}y$ is first order even if it can be written in the equivalent form $y = \sin(\frac{dy}{dx}).$
