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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.

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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.

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  • $\begingroup$ Could you please explain to me that why you have taken one y' out of diff equation ?? $\endgroup$ – Epsilon Oct 12 '18 at 5:40
  • $\begingroup$ 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 $1$. $\endgroup$ – Ng Chung Tak Oct 12 '18 at 8:05
  • $\begingroup$ Thank you for clarification $\endgroup$ – Epsilon Oct 15 '18 at 9:14

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