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