Help solving the Initial Value Problem

[middieman147]

Homework Statement

y''-4y'+4y=0 , y(1)=1 and y'(1)=1

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=e^2t
y₂=y₁

y(t)=c₁e^2t+c₂e^2t

y(1)=c₁e²+c₂e²=1 ---eq(1)y'(t)=2c₁e^2t+2c₂e^2t

...c₂=1/(2e²)-c1 ---eq(2)

sub eq(2) into eq(1) --> 1/2=1method 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 doesn't seem correct

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

y₁=c₁e^2t
y₂=c₂e^2t *t

can someone explain to me why the extra t shows up in y₂? Thanks.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

hi middieman147

(i don't understand your second method at all )

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

y₁=c₁e^2t
y₂=c₂e^2t *x

can someone explain to me why the extra x shows up in y₂?

it isn't an x, it's a t

when a root b is repeated (in the auxiliary equation), the two independent solutions are e^bt and te^bt

if it's repeated n times, the n independent solutions are e^bt te^bt … t^n-1e^bt