The solution of the initial value problem:
$y^{(4)} - 10y'' + 25y' = 0$
y(3) = -4, y'(3) = 5, y''(3) = 0, y'''(3) = 0
is of the form:
y(t) = \alpha t + \beta
Fill in the blanks with correct numbers for the values of $\alpha$ and $\beta$:
$\alpha = $
$\beta = $