#### Answer

$y= \frac{1}{3} +\frac{1}{5}t^2+\frac{1}{2}t+Ct^{-3} $

#### Work Step by Step

Rewrite the equation in the form
$y'+\frac{3}{t}y=\frac{1}{t}+t+2$
This is a linear equation and has the integrating factor as follows $$\alpha(t)= e^{\int P(t)dt}=e^{ \int \frac{3}{t} dt}=e^{3\ln t}=t^3.$$
Now the general solution is
\begin{align}
y& =\alpha^{-1}(t)\left( \int\alpha(t) Q(t)dt +C\right)\\
& =t^{-3}\left( \int (t^2+t^4+2t^3)dt+C\right)\\
& =t^{-3} \left( \frac{1}{3}t^3+\frac{1}{5}t^5+\frac{2}{4}t^4+C\right)\\
& = \frac{1}{3} +\frac{1}{5}t^2+\frac{1}{2}t+Ct^{-3} .\end{align}