Register to reply

Reduction of Order

by vipertongn
Tags: order, reduction
Share this thread:
vipertongn
#1
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
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.
Phys.Org News Partner Science news on Phys.org
Sapphire talk enlivens guesswork over iPhone 6
Geneticists offer clues to better rice, tomato crops
UConn makes 3-D copies of antique instrument parts
rock.freak667
#2
Aug19-10, 06:17 PM
HW Helper
P: 6,207
From u"e2x=0, you can divide by e2x and solve u''=0.
vipertongn
#3
Aug19-10, 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?

Mark44
#4
Aug19-10, 06:43 PM
Mentor
P: 21,216
Reduction of Order

You need initial conditions in order to solve for the constants c1 and c2.
vipertongn
#5
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
Mark44
#6
Aug19-10, 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.


Register to reply

Related Discussions
2nd order differential equation using reduction of order Calculus & Beyond Homework 7
Reduction of order (2nd order linear ODE homogeneous ODE) Calculus & Beyond Homework 10
Reduction of order Calculus & Beyond Homework 1
ODE - Reduction of order Calculus & Beyond Homework 1