$$\text{Solve the following given equation for $t > 0$.}$$
$$ \text{No need to do an actual change of variables, just use the result of Problem 34, Section 3.3}$$
$$ t^2 y^{\prime \prime} + 3 t y^{\prime} - 3y = 0 $$
$$ $$
$$ \text{characteristic equation: } r^2 - r + 3r -3 = 0 $$
$$ r^2 + 2r - 3 =0 $$
$$ \therefore r = \dfrac{-2 \pm \sqrt{16}}{2} \implies r_1 = 1 \text{ and } r_2 = - 3$$
$$y_1 = e^{x} \text{ and } y_2 = e^{- 3x} $$
$$ \therefore y\left(x\right) = c_1 e^{x} + c_2 e^{- 3x}$$
$$ x = \ln{t} $$
$$ \therefore y\left(t\right) = c_1 e^{\ln{t}} + c_2 e^{- 3 \ln{t}} $$
$$ \therefore y\left(t\right) = c_1 t + \dfrac{c_2}{t^3} $$
