Answer
$$
t^{2} y^{\prime \prime}+t y^{\prime}+y=0 , \quad t > 0
$$
The general solution of the given equation
$$
y(t) = c_{1} \cos ( \ln t)+c_{2} \sin ( \ln t)
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
t^{2} y^{\prime \prime}+t y^{\prime}+y=0 , \quad t > 0 \quad (1).
$$
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
$$
\frac{d^{2} y}{d x^{2}}+ y=0, \quad (2)
$$
We assume that $ y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+1=0,
$$
so its roots are
$$r_{1}=i ,\quad r_{2}= -i .$$
Therefore two solutions of Eq. (2) are
$$
y_{1}(x) = e^{i x}=(\cos x + i \sin x)
$$
and
$$
y_{2}(x) = e^{-ix }=(\cos x - i \sin x)
$$
Thus the general solution of the differential equation (2) is
$$
y(x) = c_{1} \cos x+c_{2} \sin x
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Back to t:
$$
y(t) = c_{1} \cos ( \ln t)+c_{2} \sin ( \ln t)
$$
