#### Answer

$$
t^{2} y^{\prime \prime}+4t y^{\prime}+2y=0 , \quad t > 0
$$
the general solution of the given differential equation is
$$
y(t) =c_{1} t^{-1} +c_{2} t^{-2}
$$

#### Work Step by Step

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