Answer
$y=\dfrac{t^3}{5}-\dfrac{12t^{-2}}{5}$
Work Step by Step
The standard form of the given equation is:
$ty'+2y=t^3$ ....(1)
The integrating factor is: $e^{\int (2/t) dt}=t^2$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [t^2y]' dt=\int t^4$
or, $t^2y=\dfrac{t^5}{5}+c$
Applying the initial conditions in the above equation, we get
$(1)(2^2)=\dfrac{(2)^5}{5}+c \implies c=-\dfrac{12}{5}$
and $t^2y=\dfrac{t^5}{5}-\dfrac{12}{5}$
Thus, the general solution is as follows:
$y=\dfrac{t^3}{5}-\dfrac{12t^{-2}}{5}$
