Answer
$$y= -x+\frac{1}{2} +\frac{40}{9}x^2 .$$
Work Step by Step
Rewrite the equation in the form
$y'-\frac{2}{x}y=1-\frac{1}{x}$.
This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ \int \frac{-2}{x}dx}=e^{-2\ln x}=\frac{1}{ x^2}$$
Now the general solution is
\begin{align}
y& =\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\
& = x^2 \left( \int ( \frac{1}{x^2}-\frac{1}{x^3})dx+C\right)\\
& =x^2 \left(- \frac{1}{x}+\frac{1}{2x^2}+C\right)\\
& = -x+\frac{1}{2} +Cx^2\end{align}
Since, $y(3/2)=9$, then $C=\frac{40}{9}$. Thus, the general solution is:
$$y= -x+\frac{1}{2} +\frac{40}{9}x^2 .$$
