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Homework Statement
[itex]y''-2y+y=xe^xlnx[/itex]
The Attempt at a Solution
I don't know what I should do here because lnx. Is it possible to solve this ODE with undetermined coefficients method? how can I solve it?
Try variation of parameters. http://en.wikipedia.org/wiki/Variation_of_parametersHomework Statement
[itex]y''-2y+y=xe^xlnx[/itex]
The Attempt at a Solution
I don't know what I should do here because lnx. Is it possible to solve this ODE with undetermined coefficients method? how can I solve it?
The problem is I don't know what should be the general form that I should consider to determine the coefficients. That's where I'm stuck.Hi AdrianZ!
Sounds like a good plan.
Did you try the undetermined coefficients method?
As for the lnx, typically you can integrate that using integration by parts.
I read the wikipedia page, I don't know what that method is, the wikipedia explanations are a bit confusing but I've already solved the ODE using the method where we take y_{p} of the form v_{1}y_{1}+v_{2}y_{2} and then we find v_{1} and v_{2} using a linear system of equations that are dependent to the derivatives of v's.Try variation of parameters. http://en.wikipedia.org/wiki/Variation_of_parameters
It is quite easy in this case.
ehild
You applied exactly the method of "variation of parameters"I read the wikipedia page, I don't know what that method is, the wikipedia explanations are a bit confusing but I've already solved the ODE using the method where we take y_{p} of the form v_{1}y_{1}+v_{2}y_{2} and then we find v_{1} and v_{2} using a linear system of equations that are dependent to the derivatives of v's.
Ln(x)has to appear in the solution anyway. You know the solution now, but pretend that you do not know, and choose a trial function according to it but with unknown constants. But I do not think they meant that.The thing is that the author has put this problem in the section of U.C method, so I'm wondering if he wants us to solve it using the undetermined coefficients method?
No, they are basically different.Hmm, as I see it both methods are basically the same.
Ah, okay.No, they are basically different.
The method of undetermined coefficients uses trial functions. There are listed such ones for different types of functions on the right hand side in relation with the roots of the characteristic equation of the homogeneous part.See attachment.
ehild
I have noticed it already and sent a PM to you about it.Hey ehild!
I've just put your explanation and attachment to good use here:
https://www.physicsforums.com/showthread.php?t=554093
Isn't that just the method of Variation of parameters?And what is this method called? :tongue:
Let be y=e^{x} Y.
Substituting into the ode, we get the equation Y"=xln(x), so Y=x^{3}(6 ln(x)-5)/36, y_{p}=e^{x}x^{3}(6 ln(x)-5)/36.
ehild