Answer
$$ y=\frac{1}{3t^3}\left(1+t^2\right)^{\frac{3}{2}}+\frac{C}{t^3}$$
Work Step by Step
Given $$t^{2} \frac{d y}{d t}+3 t y=\sqrt{1+t^{2}}, \quad t>0$$
Rewrite the equation
$$ \frac{d y}{d t}+\frac{3}{t} y=\frac{\sqrt{1+t^{2}}}{t^2} $$
Since the equation is linear and $p(t)=\frac{3}{t} $ and $q(t)=\frac{\sqrt{1+t^{2}}}{t^2}$, then
\begin{align*}
\mu &=e^{\int p(t)}dt\\
&= e^{\int (\frac{3}{t})dt}\\
&=e^{3\ln t}\\
&= t^3
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(t)y&=\int \mu(t)q(t)dt\\
t^3 y&= \int t^3 \frac{\sqrt{1+t^{2}}}{t^2} dt\ \ \\
&=\int t \sqrt{1+t^{2}} dt\ \ \\
&=\frac{1}{3}\left(1+t^2\right)^{\frac{3}{2}}+C
\end{align*}
Then
$$ y=\frac{1}{3t^3}\left(1+t^2\right)^{\frac{3}{2}}+\frac{C}{t^3}$$
