Answer
$y=x^2e^{-x}(3x-3)$
Work Step by Step
Here, we have $y'+(\dfrac{x-2}{x}) y=3x^2e^{-x}$
The integrating factor is: $I=e^{\int (\dfrac{x-2}{x}) dx}=\dfrac{e^x}{x^2}$
Now, $(\dfrac{e^x}{x^2})[y'+(\dfrac{x-2}{x}) y]=(\dfrac{e^x}{x^2})(3x^2e^{-x})$
This implies that
$\int [\dfrac{e^x}{x^2}y]' =3 \int dx $
$\implies \dfrac{e^xy}{x^2}=3x+c$
Now, apply the initial conditions.
Thus, we have $c=-3$
$\dfrac{e^xy}{x^2}=3x-3 \implies y=(x^2e^{-x})(3x-3)$
