Answer
$$
t^{2} y^{\prime \prime}-4t y^{\prime}+6 y=0 ,\quad t > 0
$$
the general solution of the given differential equation is
$$
y(t) =c_{1} t^{2} +c_{2} t^{3}
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
t^{2} y^{\prime \prime}-4t y^{\prime}+6 y=0 , \quad t > 0 \quad (1).
$$
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
$$
\frac{d^{2} y}{d x^{2}} -5 \frac{d y}{d x}+ 6 y=0, \quad (2)
$$
We assume that $ y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}-5r+6=0,
$$
so its roots are
$$
\:r_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\left(6\right)}}{2\cdot \:1}
$$
Thus the possible values of $r$ are
$$r_{1}= 2 ,\quad r_{2}=3 .$$
Therefore the general solution of the differential equation (2) is
$$
y(x) = c_{1} e^{2x}+c_{2} e^{3x}
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
Back to t:
$$
\begin{split}
y(t) & = c_{1} e^{2 \ln t}+c_{2} e^{3 \ln t} \\
& = c_{1} e^{( \ln t^{2} )}+c_{2} e^{( \ln t^{3})} \\
& =c_{1} t^{2} +c_{2} t^{3}
\end{split}
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
