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AdrianZ
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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?
AdrianZ said: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?
I like Serena said: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.
ehild said: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"AdrianZ said: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 yp of the form v1y1+v2y2 and then we find v1 and v2 using a linear system of equations that are dependent to the derivatives of v's.
AdrianZ said: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?
I like Serena said:Hmm, as I see it both methods are basically the same.
ehild said: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 like Serena said:Hey ehild!
I've just put your explanation and attachment to good use here:
https://www.physicsforums.com/showthread.php?t=554093
ehild said:And what is this method called?
Let be y=ex Y.
Substituting into the ode, we get the equation Y"=xln(x), so Y=x3(6 ln(x)-5)/36, yp=exx3(6 ln(x)-5)/36.
ehild
A non-homogenous second order ODE is one that has a non-zero term that is not multiplied by the dependent variable or its derivatives. In other words, it does not have the form of y'' + P(x)y' + Q(x)y = 0. Instead, it has the form of y'' + P(x)y' + Q(x)y = f(x), where f(x) is a non-zero function.
The general approach to solving a non-homogenous second order ODE involves finding the general solution to the associated homogeneous equation, and then using the method of variation of parameters to find a particular solution to the non-homogeneous equation. The general solution is then given by the sum of the homogeneous and particular solutions.
The general solution to the associated homogeneous equation can be found by assuming a solution of the form y(x) = e^(rx), where r is a constant to be determined. Plugging this into the homogeneous equation will give a characteristic equation, which can be solved for r. The solutions to the characteristic equation will then be used to construct the general solution y(x) = c1e^(r1x) + c2e^(r2x).
The method of variation of parameters involves finding a particular solution to the non-homogeneous equation by assuming a solution of the form y(x) = u1(x)y1(x) + u2(x)y2(x), where y1(x) and y2(x) are linearly independent solutions to the associated homogeneous equation. The functions u1(x) and u2(x) are then determined by plugging this assumed solution into the non-homogeneous equation and solving for u1(x) and u2(x).
Yes, there are two special cases for solving non-homogeneous second order ODEs: when the non-homogeneous term f(x) is a polynomial function or when it is an exponential function. In these cases, there are specific methods that can be used to find a particular solution without using the method of variation of parameters.