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Could use a little advice here, with a 2nd order ODE

by MaximumTaco
Tags: advice, order
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Oct4-04, 07:53 AM
P: 47
xy'' -(2x+1)y' + (x+1)y = (x ex)2

I know a solution - (x-1)e2x

Thus, y= ((x-1)e2x 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

(x2-x)u'' + (2x2-x+1)u' + x2u = x2

So, how do i approach this? Thanks.
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Oct4-04, 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


You already know that [tex]y_{p}(x)=(x-1)e^{2x}[/tex].

For the homogeneous equation, there are two linearly independent solutions, one of them is clearly


and using the reduction of order, the other is


So the general solution to the ode is

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.
Oct4-04, 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.

Oct5-04, 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


implies that



substituting in [tex]xy''_{2}(x)-(2x+1)y'_{2}(x)+(x+1)y_{2}(x)=0[/tex]

we get that


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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