I am trying to find the degree of the following differential equation: $$ \sqrt{\frac{d^2y}{dx^2}}+\frac{dy}{dx}=y^3. $$
I am not $100$% sure, but I know (correct me if I'm wrong) that for a differential equation to have a degree, it has to be able to be represented as a linear polynomial like the following :
$a_{0}(x)y+a_{1}(x)y'+a_{2}(x)y''+...+a_{n}(x)y^{(n)}+b(x)=0$
Therefore, I did the following reasoning :
$\sqrt{\frac{d^2y}{dx^2}}+\frac{dy}{dx}=y^3\Rightarrow \sqrt{\frac{d^2y}{dx^2}}=y^3-\frac{dy}{dx}\Rightarrow \frac{d^2y}{dx^2}=(y^3-\frac{dy}{dx})^2\Rightarrow \frac{d^2y}{dx^2}=y^6-2y^3\frac{dy}{dx}+(\frac{dy}{dx})^2\Rightarrow -y^6+(2y^3-\frac{dy}{dx})\frac{dy}{dx}+\frac{d^2y}{dx^2}=0$
After representing the differential equation in a linear polynomial form, I concluded that the degree of the differential equation is $1$. But for some reason, the term: $(2y^3-\frac{dy}{dx})\frac{dy}{dx}$, puts me off and I don't know if this is considered to be a linear term. If any of you people could help me out, that would be very kind of you.
Differential equations have both a degree and an order. The order of a differential equation is order of its highest-order derivative. The degree is the power to which the highest order derivative is raised after re-expressing the equation with integral powers. So after the squaring you performed, the highest derivative has order 2 and its power is 1. So the order is 2 and the degree is 1.
Diff Eq Terms