Solve Reduction of Order: y" - 4y' + 4y = 0

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Homework Help Overview

The discussion revolves around solving the second-order linear homogeneous differential equation y" - 4y' + 4y = 0 using the method of reduction of order, given one solution y1 = e^(2x).

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply reduction of order by substituting y2 = uy1 into the differential equation and simplifying. Some participants suggest dividing by e^(2x) to simplify the resulting equation further. Others explore the implications of u'' = 0 and discuss the integration process for finding u.

Discussion Status

Participants are actively discussing the steps involved in solving for the constants c1 and c2 after determining u. There is mention of needing initial conditions to find specific values for these constants, and some participants express confusion regarding the expected solution form.

Contextual Notes

There is a reference to a solutions manual that provides a specific solution, which raises questions about the derivation of that solution. The discussion also touches on the need for initial conditions to determine the constants in the general solution.

vipertongn
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Homework Statement



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

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^2x-4u'e^2x-8ue^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.
 
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From u"e2x=0, you can divide by e2x and solve u''=0.
 
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?
 
You need initial conditions in order to solve for the constants c1 and c2.
 
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
 
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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