Reduction of Order

Link88 · #1 $Old$ November 2nd, 2009, 05:14 PM

y''-2y'+y=0 ; With Solution y = e^x
So far I got:

let y=f(x)*v = e^xv

y'=e^xv+e^xv'

y''=e^xv+e^xv'+e^xv'+e^xv''

substituting that into y''-2y'+y=0 gets=

e^xv'' and then v''=w' we get

e^xw' not sure where to go from here though : (

Prove It · #2 $Old$ November 2nd, 2009, 05:33 PM

Quote:

Originally Posted by Link88 $View Post$

y''-2y'+y=0 ; With Solution y = e^x
So far I got:

let y=f(x)*v = e^xv

y'=e^xv+e^xv'

y''=e^xv+e^xv'+e^xv'+e^xv''

substituting that into y''-2y'+y=0 gets=

e^xv'' and then v''=w' we get

e^xw' not sure where to go from here though : (

Since this is a second order linear constant coefficient ODE, find the characteristic equation

$m^2 - 2m + 1 = 0$

$(m - 1)^2 = 0$

$m = 1$ , a repeated root.

So your solution to the DE will have the form

$y = C_1e^{x} + C_2xe^{x}$ .

You will need to use some intial conditions to find the constants.