- #1
Odyssey
- 87
- 0
Hello,
I am stuck on classifying the points with this DE...=\
xy''+(x-x^3)y'+(sin x)y=0
The solution says (sin x)/x is infinitely differentiable...so x=0 is an ordinary point?
I was taught...if P(xo)=0, then xo is a singular point. Here P(x)=x...so x=0. So, what I don't get is the "infintely differentiable" part. Does it have something to do with the convergent Tayloe Series about x=0? And what does that mean?
I got another example here...
x^2(y'')+(cos x)y'+xy=0
Here P(x)=0 so x=0 is a singular point. It's also a irregular singular point because as x->0 (cos x)/x goes to infinity. How come here x=0 is a singular point here?
Thank you.
I am stuck on classifying the points with this DE...=\
xy''+(x-x^3)y'+(sin x)y=0
The solution says (sin x)/x is infinitely differentiable...so x=0 is an ordinary point?
I was taught...if P(xo)=0, then xo is a singular point. Here P(x)=x...so x=0. So, what I don't get is the "infintely differentiable" part. Does it have something to do with the convergent Tayloe Series about x=0? And what does that mean?
I got another example here...
x^2(y'')+(cos x)y'+xy=0
Here P(x)=0 so x=0 is a singular point. It's also a irregular singular point because as x->0 (cos x)/x goes to infinity. How come here x=0 is a singular point here?
Thank you.