Second order linear equation with x and x·ln(x) as solutions

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
3 replies · 2K views
brad sue
Messages
270
Reaction score
0
Hi ,
I am stuck with the following problem:

Find a second order linear homogeneous equation having the pair as a fundamental set of solutions:
y1(x)=x , y2(x)=x*ln(x).

My problem here is that I don't have the exponential form for the proposed solutions.

Thank you for your help

B.
 

Discussion

Physics news on Phys.org
You don't need an exponential form. Can you find a (single) linear combination of the function and its first and second derivatives that add to zero - for both functions?
 
Tide said:
You don't need an exponential form. Can you find a (single) linear combination of the function and its first and second derivatives that add to zero - for both functions?

Ok you mean :
y1=x ----> (y1)'=1 -------> (y1)"=0
it gives: x*y' -y=0

y2(x)=x*ln(x) ----> (y2)'=1+ln(x) -----> (y2)"=1/x
it gives y*y"-(y'-1)=0

The second equation seems to be the good one since it is a second degree?
Am I right?
 
I don't think there is any need to go nonlinear. Try this:

x y'' - x y' + y = 0