Final answer:
Upon verification, y1(t) = -1 is not a solution to the given differential equation as its substitution results in a non-zero value. However, y2(t) = t^-1 is a solution as its substitution satisfies the equation.
Step-by-step explanation:
To verify that y1(t) = -1 and y2(t) = t-1 are solutions to the differential equation t2y'' - 2y = 0, we need to substitute these functions into the equation and see if the left-hand side equals zero for both functions.
Verification for y1(t) = -1:
Since y1(t) is a constant, its derivatives are zero:
y1'(t) = 0 and y1''(t) = 0.
Substitute into the equation: t2(0) - 2(-1) = 0 which simplifies to 0 + 2 = 2, hence this is not equal to zero, and y1(t) = -1 is not a solution to the differential equation.
Verification for y2(t) = t-1:
Differentiate y2(t):
y2'(t) = -t-2
Differentiate again:
y2''(t) = 2t-3
Substitute into the equation: t2(2t-3) - 2(t-1) = 2 - 2 = 0, hence y2(t) does satisfy the differential equation.