# Homework Help: Help solving the Initial Value Problem

1. Apr 1, 2012

### middieman147

1. The problem statement, all variables and given/known data
y''-4y'+4y=0 , y(1)=1 and y'(1)=1

3. The attempt at a solution

Auxiliary equation: r2-4r+4=0

I tried factoring 2 different ways:
(r-2)2=0

r=2,r=2

y1=e2t
y2=y1

y(t)=c1e2t+c2e2t

y(1)=c1e2+c2e2=1 ---eq(1)

y'(t)=2c1e2t+2c2e2t

...c2=1/(2e2)-c1 ---eq(2)

sub eq(2) into eq(1) --> 1/2=1

method 2:

r(r-4)=-4

r=-4, r=0

y(1)=c1e-4+c2=1 ---eq(1)

y'(1)=c1=-1/(4e-4)

sub eq(2) into eq(1) --> c2=5/4

solution equation:

y(t)=1/4(-1+5)
=1..... which doesnt seem correct

i checked the solutions on wolframalpha and it solves it using the first method, but ends up with:

y1=c1e2t
y2=c2e2t *t

can someone explain to me why the extra t shows up in y2? Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Apr 1, 2012
2. Apr 1, 2012

### tiny-tim

hi middieman147

(i don't understand your second method at all )
it isn't an x, it's a t

when a root b is repeated (in the auxiliary equation), the two independent solutions are ebt and tebt

if it's repeated n times, the n independent solutions are ebt tebt … tn-1ebt