How are order and degree defined for this DE? (cos(y'') + xy' = 0)

1. Aug 9, 2007

maverick280857

Hi everyone

Consider the differential equation

$$\cos(y'') + xy' = 0$$

How do you determine the order and degree of such a DE?

One way is to write

$$y'' = \cos^{-1}(-xy')$$

and say that the order is 2 degree is 1.

But if I do not use the inverse cosine, and observe that the first member on the left hand side is a power series in $y''$, then the order is still 2, but the degree is not defined. What is the resolution to this problem?

PS--This is not homework.

2. Aug 9, 2007

HallsofIvy

Staff Emeritus
That is a second order equation because y" is the highest derivative.

It does NOT HAVE a degree because it is not polynomial.

I'm not sure why you say that $y''= cos^{-1}(-xy')$ has degree one. Is it because y'' is to first power? You can always solve a differential equation for the highest derivative so that way of looking at "degree" reduces any equation to first degree!

3. Aug 9, 2007

Kummer

As mentioned about it does not have a degree.

One thing that physicists and engineers do in such a situation is linearize the differential equation. Since $$\cos (y'')\approx 1$$ for small $$y''$$. And so we get,
$$1+xy'=0$$.

But we need to be careful about the accuray.

4. Aug 9, 2007

mathwonk

it looks second order and not being polynomial in the derivatives has no finite degree.

5. Aug 18, 2007

Matthew Rodman

Yes, but it is effectively a first order equation, as all you have to do is let $$y^{\prime} = u(x)$$.

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