Someone help me with this differential equation please!!!!?

Question

y"-8y'+16y=0, y(0)=0,y'(0)=1 Ahhh!

Answer 1

Characteristic equation:

s² - 8s + 16 = 0
(s-4)² = 0
s = 4 (double root)

y = Ae^(4x) + Bx.e^(4x)
y(0) = 0
A+0 = 0
A = 0

y = Bx.e^(4x)
y' = Be^(4x) + 4Bx.e^(4x)
y'(0) = B = 1

y = x.e^4(x)

Answer 2

y"-8y'+16y=0
use H.D.E.C.C. method
Polinomy: P(x)=r^2-8r+16
think out r values:
r1=r2=4
general solution:
y(t)=A·e^(4·t)+B·t·e^(4·t)
think out A an d B values by initials conditions:
y(0)=0
y'(0)=1
obtain
A=0, B=1
replace in y(t)
y(t)=t·e^(4·t)

uffffffff!

good luck,

Answer 3

r^2-8r+16=0 so r = 4(double root
general solution
y=Ae^4x +Bt e^4x
Putting in initial conditions
0=A
y´= +B( e^4x -4xe^4x)
1=B(1) so particular solution y = x*e^4x