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Reduction of Order 
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#1
Aug1910, 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 y2=uy=ue^2x y2'=u'e^2x+2ue^2x y2"=u"e^2x+4u'e^2x+4ue^2x I then substitute that into the original equation to get u"e^2x+4u'e^2x+4ue^2x4u'e^2x8ue^2x+ue^2x=0 simplify to get u"e^2x=0 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. 


#2
Aug1910, 06:17 PM

HW Helper
P: 6,207

From u"e^{2x}=0, you can divide by e^{2x} and solve u''=0.



#3
Aug1910, 06:38 PM

P: 98

ahh...so then u"=0 makes u'=c and then later u=xc1+c2 and
y2=uy1 y2=xc1*e^2x+c2*e^2x but what then? how do I solve for c1 and c2? 


#4
Aug1910, 06:43 PM

Mentor
P: 21,216

Reduction of Order
You need initial conditions in order to solve for the constants c1 and c2.



#5
Aug1910, 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 


#6
Aug1910, 07:36 PM

Mentor
P: 21,216

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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