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What should be the degree of this differential equation 1 or 2 $$\frac{d^2x}{dy^2}+\sqrt{1+\left(\frac{dx}{dy}\right)^3}=0$$

As per definition: the degree of a differential equation is the power of highest differential coefficient, which appearing in the given equation, when the differential coefficients are free from radicals and fractions. So I think I should remove the radical in the equation above and the degree will be 2. Is it correct ?

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  • $\begingroup$ You might want to edit. I'm not seeing "the equation above". $\endgroup$ – bob.sacamento Dec 3 '15 at 17:15
  • $\begingroup$ Its given in the image $\endgroup$ – Onix Dec 3 '15 at 17:16
  • $\begingroup$ If you mean the order, it's a second order differential equation. $\endgroup$ – gerd Dec 3 '15 at 17:23
  • $\begingroup$ Degree is different to Order as he defined in the question. $\endgroup$ – Ian Miller Dec 3 '15 at 17:25
  • $\begingroup$ I think that you mean it's order. Its ODE classification: It is a second- order nonlinear ordinary differential equation $\endgroup$ – Jan Dec 3 '15 at 17:43
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A better definition is the power of the highest order derivative, when the equation is polynomial in the derivatives.

Hence

$$\left(\frac{d^2y}{dx^2}\right)^2=1+\left(\frac{dy}{dx}\right)^3$$

is of the second order and second degree.

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According the definition in the question the degree must be 2. Some textbooks use the term "coefficient differentials" meaning "derivative". In my view, kind of definition given in another answer by Yves Daoust is more suitable.

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