# 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

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 $$y^{\prime} = u(x)$$.