Determine the values of r for which the given differential equation has solutions of the form y = tr for t > 0. (Enter your answers as a comma-separated list.) t2y'' + 6ty' + 6y = 0

r = [ - 2, - 3 ]

Step-by-step explanation:

for

y = t^r

then

y' = r*t^ (r-1)

y'' = r * (r-1) * t^ (r-2)

thus if

t²y'' + 6t*y' + 6y = 0

r * (r-1) * t^ (r-2) * t² + 6*t*r*t^ (r-1) + 6*t^r = 0

t^r * [ r (r-1) + 6*r + 6 ] = 0

since t>0, then [ r (r-1) + 6*r + 6 ] = 0

r (r-1) + 6*r + 6 = 0

r² - r + 6*r + 6 = 0

r² + 5*r + 6 = 0

r₁, r₂ = [-5 ±√ (5² - 4*1*6) ]/2*1

r₁ = (-5+1) / 2 = - 2

r₂ = (-5-1) / 2 = - 3

then r = [ - 2, - 3 ]