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I understand why we define the order of differential equations but why do we need to define the degree of the equation? I don't understand how it is important for differential equations. Also, if degree is defined as the power of the highest derivative, then, for example, consider $$\frac{d^2 y}{dx^2}+\sin\left[\frac{dy}{dx}\right]+y=0$$ Here we don't define the degree of the equation because one of the derivative terms is present inside a trigonometric function. But if we write the sin term in its polynomial form, then even though the highest power of first derivative approaches infinity, the power of the highest derivative is still 1. Then why don't we define the degree in this case?

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marked as duplicate by Nosrati, Arnaud Mortier, Claude Leibovici, José Carlos Santos, Namaste Jul 24 '18 at 11:35

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An explicit differential equation $$y^{(n)}=f(x,y,y',...,y^{(n-1)})$$ defines a ($x$-dependent) vector field $$v=(y',y'',...,f(...))$$ on the state space with coordinates $$u=(y,y',...,y^{(n-1)})$$ so that $$u'=v.$$ That it is explicit could also be describes as the highest derivative having degree 1.


In an implicit ODE $$0=F(x,y,y',...,y^{(n)})$$ there might be several directions, that is solutions for $y^{(n)}$ of that equation, that satisfy the DE in a given point $(x,y,...,y^{(n-1)})$. If the highest derivative occurs as polynomial $$F=c_d(y^{(n)})^d+...+c_1y^{(n)}+c_0,$$ where its coefficients $c_k=c_k(x,y,...,y^{(n-1)})$ may depend on the independent variable and lower order derivatives, then the degree $d$ of this polynomial bounds the number of directions in that point.


As one usually considers real ODE with real solutions, this bound $d$ on directions that a solution may follow in a given point may be pessimistic, as the polynomial may have many complex roots.

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  • $\begingroup$ Uhh.. I guess this is too advanced for me; I am still in high school. $\endgroup$ – Nischay Jul 23 '18 at 16:06

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