
#1
Oct404, 07:53 AM

P: 47

xy'' (2x+1)y' + (x+1)y = (x e^{x})^{2}
I know a solution  (x1)e^{2x} Thus, y= ((x1)e^{2x} u(x)) Now, i know how to do the whole reduction of order thing, but when i find y' and y'' and substitute, the u(x) term doesn't cancel out so this doesn't work (x^{2}x)u'' + (2x^{2}x+1)u' + x^{2}u = x^{2} So, how do i approach this? Thanks. 



#2
Oct404, 06:10 PM

P: 120

You know that the solution can be written in the form [tex]y(x)=y_{p}(x)+y_{h}(x)[/tex] where [tex]y_{p}(x)[/tex] is the particular solution and [tex]y_{h}(x)[/tex] is the solution to the homogeneous equation
[tex]xy_{h}''(x)(2x+1)y_{h}'(x)+(x+1)y_{h}(x)=0[/tex]. You already know that [tex]y_{p}(x)=(x1)e^{2x}[/tex]. For the homogeneous equation, there are two linearly independent solutions, one of them is clearly [tex]y_{h_{1}}(x)=c_{1}e^{x}[/tex] and using the reduction of order, the other is [tex]y_{h_{2}}(x)=c_{2}\frac{x^2}{2}e^{x}[/tex]. So the general solution to the ode is [tex]y(x)=c_{1}e^{x}+c_{2}\frac{x^2}{2}e^{x}+(x1)e^{2x}[/tex] If you need further justification, use frobenius method for the homogenous equation to calculate [tex]y_{h_{1}}(x)[/tex], reduction of order for [tex]y_{h_{2}}(x)[/tex], and variation of parameters for [tex]y_{p}(x)[/tex]. That way the problem is fully justified. ps. i deleted my first post as it was wrong and confusing. 



#3
Oct404, 08:05 PM

P: 47

I understand the first technique, but i'm not familiar with the Frobenius method, i looked it up on Mathworld and it looks really complex
With your other post, i can't quite follow how to get the second linearly independant solution. Thanks. 



#4
Oct504, 01:45 AM

P: 120

Could use a little advice here, with a 2nd order ODE
Ignore the first post... is nonsence, thats why i deleted it.
Given [tex]y_{1}(x)=e^{x}[/tex] using reduction of order to get the second solution [tex]y_{2}(x)=v(x)e^{x}[/tex] implies that [tex]y'_{2}(x)=[v'(x)+v(x)]e^{x}[/tex] [tex]y''_{2}(x)=[v''(x)+2v'(x)+v(x)]e^{x}[/tex] substituting in [tex]xy''_{2}(x)(2x+1)y'_{2}(x)+(x+1)y_{2}(x)=0[/tex] we get that [tex]xv''(x)v'(x)=0[/tex] do i need to go further? ps. Frobenius method is used only if you want to *construct* the first solution [tex]y_{1}=c_{1}e^{x}[/tex]. Just so you know that it didnt came from divine inspiration, you dont need to use it though. 


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