Ordinary Differential Equations Problem.?

Question

determine the values of r for which the given defferential equation has solutions of the form y = t^r for t > 0.

t^2*y'' - 4ty' + 4y = 0

a step by step would be amazing but i know its difficult to do on a computer.

Answer 1

y'=rt^(r-1)

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

So,

t^2y''-4ty'+4y=r(r-1)t^r-4rt^r+4t^r=0

Divide by t^r (since t>0) to get

r(r-1)-4r+4=0

r^2-5r+4=0

(r-4)(r-1)=0

r=4

r=1