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Can anyone explain me that why the 'Degree' of the differential equation $\frac{d^{3}y}{dx^{3}}+\log(\frac{dy}{dx})=2$ is undefined whereas the 'Order' is defined and the 'Order' of this differential equation is '$3$'. In my class XIITh book(R.S. AGGARWAL) I have read that generally the 'Degree' of the differential equations involving one of the terms like $e^{\frac{dy}{dx}}$, $\log(\frac{dy}{dx})$, $\sin(\frac{dy}{dx})$, $\cos(\frac{dy}{dx})$,etc. are not defined. For e.g.- i) the 'Order' and 'Degree' of the differential equation $\frac{d^{3}y}{dx^{3}}+3\frac{dy}{dx}=4$ are '$3$' and '$1$' respectively. Similarly ii) the 'Order' and 'Degree' of the differential equation $(\frac{d^{2}y}{dx^{2}})^{2}+4y=2$ are '$2$' and '$2$' respectively. But I can't understand the actual logic why the 'Degree' of the differential equation $\frac{d^{3}y}{dx^{3}}+\log(\frac{dy}{dx})=2$ is undefined and on the other hand the 'Order' is defined. Similarly the 'Order' of the differential equation $(\frac{d^{2}y}{dx^{2}})^{2}+3\sin(\frac{dy}{dx})=4y$ is defined and the 'Degree' is undefined. The 'Order' is '$2$'. Also I have asked this question to the professors of my University but no can give me the actual logic. Almost $90$% of the professors are saying to me that it is a definition. So, we must follow the definition. Their opinion is that one shouldn't ask anything about definition. But still please anyone explain me the actual logic. I shall be highly obliged if anyone can help me out with this question.

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The order of an ODE is the highest order derivative that occurs.

The degree of an ODE is the power to which the highest order derivative is raised.

This definition of degree assumes that there is such a power. If you consider the Taylor series $$\log(1+ x) = x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4 +\cdots$$ then you can imagine it contains arbitrarily high powers, so there is no 'highest power'. However, that isn't rigorous. Simply put, "degree" is only defined if the highest order derivative appears explicitly with an exponent and not inside another function like that.

See this answer.

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  • $\begingroup$ But in my question why I will take into consider $\log(\frac{dy}{dx})$? In the differential equation $\frac{d^{3}y}{dx^{3}}+\log(\frac{dy}{dx})=2$ the 'Order' is $3$. And in your expansion of $\log(x)$ here $x$ implies $\frac{dy}{dx}$. But 'Degree' is 'raised to the power of highest order derivative'. $\endgroup$ Dec 4, 2023 at 16:07
  • $\begingroup$ @user619894 : Yes, corrected, thanks. I used that series because $\log x$ has no Taylor expansion about $x=0$. As I said, this isn't rigorous - it just suggests a plausible reason. $\endgroup$
    – MPW
    Dec 4, 2023 at 17:10
  • $\begingroup$ @SyamaprasadChakrabarti : Ah, I understand what you mean. The degree would be 1 according to what I have said. Possibly "degree" is only applicable to equations which contain only polynomial functions of the derivatives of $y$ (including $y$ itself). This may make sense in interpreting "degree" in a broader context. $\endgroup$
    – MPW
    Dec 4, 2023 at 17:24
  • $\begingroup$ For sure, I agree. But, if we switch variables, the problem starts to be different. For this case, I had an explicit solution like $x'=f(y')$ $\endgroup$ Dec 5, 2023 at 1:59

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