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## Homework Statement

[itex]t^2y" + 5ty' + 4y = 0[/itex]

Possible solutions:

[itex]y_1(t) = t^{-2}[/itex] and [itex]y_2(t) = t^{-2} lnt[/itex]

## Homework Equations

## The Attempt at a Solution

I was able to verify that y_1 was a solution, by the substituting the function, and its derivatives, into the differential equation, which resulted in the identity 0 = 0, this being true for every t.

However, for the second function, my work yielded the expression [itex]16t^2 lnt +t^3 + 7t^2 = 0[/itex]. My question is, does this function satisfy the differential equation only if, say, [itex]t_o[/itex] is the value that results in the left side of equation being zero?