Answer
$$
t^{2} y^{\prime \prime}+3 t y^{\prime}+1.25 y=0 , \quad t > 0
$$
the general solution of the given differential equation is
$$
y(t) =t ^{-1} [ c_{1} \cos ( \frac{1}{2}\ln t)+ c_{2}\sin ( \frac{1}{2} \ln t) ]
$$
where $ c_{1} $and $c_{2}$ are arbitrary constants.
Work Step by Step
$$
t^{2} y^{\prime \prime}+3 t y^{\prime}+1.25 y=0 , \quad t > 0 \quad (1).
$$
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
$$
\frac{d^{2} y}{d x^{2}}+ 2\frac{d y}{d x}+ 1.25 y=0, \quad (2)
$$
We assume that $ y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation
$$
r^{2}+2r+1.25=100r^2+200r+125=0,
$$
so its roots are
$$
\:r_{1,\:2}=\frac{-200\pm \sqrt{200^2-4\cdot \:100\cdot \:125}}{2\cdot \:100}\quad
$$
Thus the possible values of $r$ are
$$r_{1}= -1+i\frac{1}{2} ,\quad r_{2}=-1-i\frac{1}{2} .$$
Therefore two solutions of Eq. (1) are
$$
y_{1}(x) = e^{(-1+i\frac{1}{2} )x}= e^{(-x)} (\cos \frac{1}{2}x + i \sin \frac{1}{2} x )
$$
and
$$
y_{2}(x) = e^{(-1-i\frac{1}{2} )x}= e^{(-x)} (\cos \frac{1}{2}x - i \sin \frac{1}{2} x )
$$
Thus the general solution of the differential equation is
$$
y(x) = e^{(-x)} ( c_{1} \cos \frac{1}{2}x+ c_{2}\sin \frac{1}{2} x ) \quad\quad\quad (2)
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
Back to t:
$$
\begin{split}
y(t) & = e^{(-\ln t)} ( c_{1} \cos \frac{1}{2}\ln t+ c_{2}\sin \frac{1}{2} \ln t ) \\
& = e^{(\ln t ^{-1})} ( c_{1} \cos \frac{1}{2}\ln t+ c_{2}\sin \frac{1}{2} \ln t ) \\
& =t ^{-1} [ c_{1} \cos ( \frac{1}{2}\ln t)+ c_{2}\sin ( \frac{1}{2} \ln t) ]
\end{split}
$$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.
