General Solution to Linear, Homogenius, Second-Order Differential Equation

  • Thread starter nivekious
  • Start date
  • #1
5
0
I need to solve d/dx(e^(-(x^2))*(du/dx))=u+x*(du/dx)
I get e^(-(x^2))*u''-2xe^(-(x^2))*u'=u+x*u'
u''-2x*u'=e^(x^2)*u+xe^(x^2)*u'
u''+(2x-xe^(x^2))*u'-e^(x^2)*u=0
but I have no idea how to approach the problem from here. Can somebody please help?
 

Answers and Replies

  • #2
1,796
53
I need to solve d/dx(e^(-(x^2))*(du/dx))=u+x*(du/dx)
I get e^(-(x^2))*u''-2xe^(-(x^2))*u'=u+x*u'
u''-2x*u'=e^(x^2)*u+xe^(x^2)*u'
u''+(2x-xe^(x^2))*u'-e^(x^2)*u=0
but I have no idea how to approach the problem from here. Can somebody please help?
I can suggest a route to study it. First, get a DE text that deals with second-order DEs with variable coefficients. I use "Intermediate Differential Equations" by Rainville. There are some simple substitutions to try with these types of equations, putting it in it's normal form [itex]v''+Iv=0[/itex] and sometimes that results in a constant-coefficient equation. Some other substitution method to get a constant-coefficient equation. I tried those quickly and did not seem to do it. But you may want to try yourself. My next step would be to turn to Mathematica in desperation using the DSolve command. I tried that and Mathematica can't solve it. My next step, even more desperate but not too uncommon is to use power series which of course in this case will involve Cauchy-products but that's ok according to Rainville. If I had to solve it analytically, and I could think of nothing else, I'd do it with power series.
 
  • #3
798
34
General Solution to Linear, Homogenius, Second-Order Differential Equation
I don't think that the ODE is on the homogenius kind. It doesn't matter since direct integration directly leads to a first order ODE :
 

Attachments

  • #4
1,796
53
. . . schooled. Very nice though. :)
 

Related Threads on General Solution to Linear, Homogenius, Second-Order Differential Equation

Replies
1
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
1
Views
5K
  • Last Post
Replies
9
Views
942
Replies
4
Views
2K
Top