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

[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.

 

[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!

 

[Kummer]

Hi everyone

Consider the differential equation

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

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

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.

 

[mathwonk]

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

 

[Matthew Rodman]

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