- #1

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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?

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- Thread starter nivekious
- Start date

- #1

- 5

- 0

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?

- #2

- 1,800

- 53

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

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- #4

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. . . schooled. Very nice though. :)

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