How to derive Euler DE for n=0

  • Thread starter yungman
  • Start date
  • #1
5,571
200
[itex] x^2 y'' + x y' + n^2 y = 0 \;[/itex] Is Euler equation and solution is [itex] y=x^m[/itex]

I understand the three cases with different solution. But my question is if n=0.

If I use [itex] y=x^m \;\Rightarrow\; m(m-1)+m=0 \;\Rightarrow m^2 = 0 \;\Rightarrow m=0[/itex]

That would not work. I know the answer is [itex] y=C_1 ln(x) + C_2 [/itex]

Can anyone show me or give me hint how to derive this, I can't find it in my book.

Thanks
 

Answers and Replies

  • #2
5,571
200
I got it, I tried to delete this post and can't. Moderator, please delete the entire thread for me.

Thanks
 
  • #3
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,542
582
Just as a general rule of thumb, any time you get a solution of x0 that isn't really a solution, you should check to see if ln(x) is a solution. ln(x) differentiates with respect to the power rule in the way that x0 is "supposed to" (i.e. give x-1)
 
  • #4
5,571
200
Just as a general rule of thumb, any time you get a solution of x0 that isn't really a solution, you should check to see if ln(x) is a solution. ln(x) differentiates with respect to the power rule in the way that x0 is "supposed to" (i.e. give x-1)

Yeh, right after I posted, I remember getting the second solution by reduction of order!!!
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
964
Note, by the way, that the euler type equation, [itex]a_nx^n y^{(n)}+ \cdot\cdot\cdot+ a_1x y'+ a_0y= f(x)[/itex]
can be converted to an equation with constant coefficients by substituting t= ln(x) as independent variable.

Since, if m is a double root of the characteristic equation for an equation with constant coefficients, the solution involves both [itex]e^{mt}[/itex] and [itex]te^{mt}[/itex], solutions for the Euler type equation will involve [itex]e^{mln(x)}= e^{x^m}= x^m[/itex] and [itex]ln(x)x^m[/itex].
 

Related Threads on How to derive Euler DE for n=0

Replies
3
Views
7K
  • Last Post
Replies
22
Views
4K
Replies
1
Views
3K
Replies
5
Views
1K
  • Last Post
Replies
1
Views
15K
  • Last Post
Replies
16
Views
4K
  • Last Post
Replies
10
Views
2K
Top