Answer
$$
t^{2} \frac{d^{2} y}{d t^{2}}+\alpha t \frac{d y}{d t}+\beta y=0, \quad t>0
$$
Let $ x = \ln t$ . Then we have
$$
\frac{d y}{d t}=\frac{d y}{d x} \frac{d x}{d t}=\frac{1}{t} \frac{d y}{d x}
$$
$$
\frac{d^{2} y}{d t^{2}}=\frac{d}{d t}\left(\frac{1}{t} \frac{d y}{d x}\right)=\frac{-1}{t^{2}} \frac{d y}{d x}+\frac{1}{t} \frac{d}{d t}\left(\frac{d y}{d x}\right)=\frac{1}{t^{2}}\left(-\frac{d y}{d x}+\frac{d^{2} y}{d x^{2}}\right)
$$
So if we use $x$ instead of $t$ as the variable, the equation (with unknown $y $ and variable $ x$) reads
$$
\frac{d^{2} y}{d x^{2}}+ (\alpha -1)\frac{d y}{d x}+ \beta y=0,
$$
Work Step by Step
$$
t^{2} \frac{d^{2} y}{d t^{2}}+\alpha t \frac{d y}{d t}+\beta y=0, \quad t>0 \quad (1)
$$
Let $ x = \ln t$ . Then we have
$$
\frac{d y}{d t}=\frac{d y}{d x} \frac{d x}{d t}=\frac{1}{t} \frac{d y}{d x}
$$
$$
\frac{d^{2} y}{d t^{2}}=\frac{d}{d t}\left(\frac{1}{t} \frac{d y}{d x}\right)=\frac{-1}{t^{2}} \frac{d y}{d x}+\frac{1}{t} \frac{d}{d t}\left(\frac{d y}{d x}\right)=\frac{1}{t^{2}}\left(-\frac{d y}{d x}+\frac{d^{2} y}{d x^{2}}\right)
$$
So if we use $x$ instead of $t$ as the variable, the equation (with unknown $y $ and variable $ x$) reads
$$
\frac{d^{2} y}{d x^{2}}+ (\alpha -1)\frac{d y}{d x}+ \beta y=0, \quad (2)
$$
Observe that Eq. (2) has constant coefficients. If $ y_{1}(x) $ and $y_{2}(x) $ form a fundamental set of solutions of Eq. (2), then $ y_{1}( \ln t) $ and $ y_{2}( \ln t) $ form a fundamental set of solutions of Eq. (1).
