Answer
$x=\dfrac{2}{y}+\dfrac{4}{y^2}+\dfrac{4}{y^3}-\dfrac{4e^{y+1}}{y^3}$
Work Step by Step
The standard form of the given equation is:
$x'+(\dfrac{3-y}{y})x+\dfrac{2}{y}=0$ ....(1)
The integrating factor is: $e^{\int (\dfrac{3-y}{y}) dx}=y^3e^{-y}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [y^3e^{-y} x]' =\int (-\dfrac{2}{y})y^3e^{-y} dy $
or, $y^3e^{-y} x=2y^2e^{-4}+4ye^{-y}+4e^{-y}+c$
Applying the initial conditions, we get $c=-4e$
Thus, we have : $y^3e^{-y} x=2y^2e^{-4}+4ye^{-y}+4e^{-y}-4e$
Hence, $x=\dfrac{2}{y}+\dfrac{4}{y^2}+\dfrac{4}{y^3}-\dfrac{4e^{y+1}}{y^3}$
