Find all REAL solutions, for x0 of:
x^3y'''+3x^2y''+xy'+y=0

This is Differential Equations but I have no idea how to go about solving this. Help?

Question

Find all REAL solutions, for x>0 of:
x^3y'''+3x^2y''+xy'+y=0

This is Differential Equations but I have no idea how to go about solving this. Help?

tkhunny · Answer

You could take a wild guess.  See what y = ln(x) does.  Maybe it will lead somewhere interesting.

anonymous · Answer

There isn't a process to follow? I just need to guess what y could be?

anonymous · Answer

I think I follow what you are saying with the ln(x) but shouldn't there be three solutions?

dumbcow · Answer

here is solution at least: http://www.wolframalpha.com/input/?i=y%27%27%27+%2B+%283%2Fx%29y%27%27+%2B%281%2Fx%5E2%29y%27+%2B+%281%2Fx%5E3%29y+%3D+0 sorry i am having trouble figuring out how to solve it though. here is a good reference to give you some ideas http://tutorial.math.lamar.edu/Classes/DE/HOUndeterminedCoeff.aspx

tkhunny · Answer

Seriously, many of the fancy methods you will encounter are just well-trained ways to make appropriate guesses.  If something rational pops into your head, why not use it?  This is the intent of my comment.  Your own experience and judgment will help you.

tkhunny · Answer

I would be tempted to try ln(x) because it's sign-alternating derivatives seem to fit the pattern.

I am tempted to add x, x^2, and x^3, since those MAY contribute to other solutions.

I might be tempted to try sin(x) and cos(x), and possibly xsin(x) etc.  Three derivatives get us back to where we started.

anonymous · Answer

Thank you both. I'm still a bit confused but maybe I'll get it lol.