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


by MaximumTaco
Tags: advice, order
MaximumTaco
MaximumTaco is offline
#1
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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ReyChiquito
ReyChiquito is offline
#2
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

[tex]xy_{h}''(x)-(2x+1)y_{h}'(x)+(x+1)y_{h}(x)=0[/tex].

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

[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}+(x-1)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.
MaximumTaco
MaximumTaco is offline
#3
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.

ReyChiquito
ReyChiquito is offline
#4
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

[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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