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Reduction of Order

by vipertongn
Tags: order, reduction
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Aug19-10, 05:18 PM
P: 98
1. The problem statement, all variables and given/known data

solve y"-4y'+4y=0 y1=e^(2x) using reduction of order

3. The attempt at a solution

I then substitute that into the original equation to get


simplify to get

from here I do not know what to do...I do know the answer is suppose to be xe^2x, but I don't know how that is done.
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Aug19-10, 06:17 PM
HW Helper
P: 6,202
From u"e2x=0, you can divide by e2x and solve u''=0.
Aug19-10, 06:38 PM
P: 98 then u"=0 makes u'=c and then later u=xc1+c2 and

but what then? how do I solve for c1 and c2?

Aug19-10, 06:43 PM
P: 21,284
Reduction of Order

You need initial conditions in order to solve for the constants c1 and c2.
Aug19-10, 06:49 PM
P: 98
however, in my solutions manual it says the solution comes out to be xe^2x, and I have no idea how that came to be. except for the use of this equation
y2=y1S e^(-SP(x)dx)/y1^2 dx
Aug19-10, 07:36 PM
P: 21,284
The general solution of your diff. equation is y = c1e^(2x) + c2xe^(2), for any values of c1 and c2. The simplest pair of linearly independent solutions is the pair with c1 = c2 = 1, so maybe they just arbitrarily chose that one.

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