# Finding order and degree of a differential equation

The question was

Find the sum of degree and order of the given DE (differential equation)$$\frac{d}{dx} \left(\frac{dy}{dx}\right)^3=0$$

So we have that $$\left(\frac{dy}{dx}\right)^3=c$$ whee c is some constant. For knowing the order and degree of a DE, it should not contain any arbitrary constant. $$3\left(\frac{dy}{dx}\right)^2\left(\frac{d^2y}{dx^2}\right)=0$$ But this is not a polynomial form so degree shold not be defined. But the answer says that degree is 3 and order is 1.

Let $$f(x,y,y',...,y^{(n)})=0$$, where $$f(.)$$ is a polynomial for multi-variables. "Order" refers to highest order of derivatives, $$n$$ in this case. "Degree" refers to the highest power of $$y^{(n)}$$ in $$f(.)$$.
For $$\frac{d}{dx} \left( \frac{dy}{dx} \right)^{3} =0$$,
$$3\left( \frac{dy}{dx} \right)^{2} y''=0$$ which has one repeated root $$y'=0$$.
Then $$y'y''=0$$, this is second order degree $$1$$ nonlinear ODE.
• First of all, don't mess up with the characteristic equation of an ODE. For example, $y''-2y'+y=0$, $$y(x)=ae^x+b\color{red}{xe^x}$$ in this case, the repeated root does alter the form of the general solution. However in $(y')^2=0$, the repeating factor $y'$ doesn't give rise to an extra form in the solution. That's why the degree still being $1$. – Ng Chung Tak Oct 12 '18 at 8:05