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 ]

1010: 3030

is it i think