Answer
$$
t^{2} y^{\prime \prime}+7 t y^{\prime}+10 y=0 , \quad t > 0
$$
the general solution of the given differential equation is
$$
y(t) = t^{-3} ( c_{1} \cos (\ln t)+ c_{2} \sin ( \ln t) )
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
t^{2} y^{\prime \prime}+7 t y^{\prime}+10 y=0 , \quad t > 0 \quad (1).
$$
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
$$
\frac{d^{2} y}{d x^{2}}+6\frac{d y}{d x}+ 10 y=0, \quad (2)
$$
We assume that $ y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+6r+10=0,
$$
so its roots are
$$
\:r_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \:10}}{2\cdot \:1}\quad
$$
Thus the possible values of $r$ are
$$r_{1}= -3+i ,\quad r_{2}=-3- i .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(x) = e^{(-3+i )x}= e^{-3x} ( \cos x + i \sin x )
$$
and
$$
y_{2}(x) = e^{(-3-i )x}= e^{-3x} ( \cos x - i \sin x )
$$
Thus the general solution of the differential equation (2) is
$$
y(x) = e^{(-3x)} ( c_{1} \cos x+ c_{2} \sin x )
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Back to t:
$$
\begin{split}
y(t) & = e^{(-3\ln t)} ( c_{1} \cos (\ln t)+ c_{2} \sin ( \ln t) ) \\
& = e^{(\ln t^{-3})} ( c_{1} \cos (\ln t) + c_{2} \sin ( \ln t )) \\
& = t^{-3} ( c_{1} \cos( \ln t)+ c_{2} \sin (\ln t) )
\end{split}
$$
where $ c_{1} $and $c_{2} $ are arbitrary constants.
