Find General Solution of First Order DE. 
Find general solution of
  $$y'+\frac{2y}{x}=\frac{3}{x^2}$$
  and find solution for which
  $$y(2)=-1$$
  I know
  $$xy'+\frac{r(x)y}{x}=3r(x)x$$
  $$(ry')=ry'+r'y$$$$r'=\frac{r}{x},\frac{r'}{r}=\frac{1}{x},ln(r)=ln(x),r=x$$$$\frac{xy'}{2}+y=\frac{3}{2x}$$
  $$\frac{(xy)'}{2}=\frac{3}{2x}$$$$\frac{xy}{2}=\int \frac{3}{x}dx=3ln|x|+c$$
  but where to from here?


It is a non-homogeneous linear ODE and IVP. Find the general solution in two steps:
1) Solve the homogeneous equation:
$$y'+\frac{2y}{x}=0 \Rightarrow\frac{1}{y}dy=-\frac{2}{x}dx \Rightarrow y=\frac{C}{x^2}.$$
2) Assume $y=\frac{C(x)}{x^2}:$
$$y'+\frac{2y}{x}=\frac{3}{x^2} \Rightarrow \frac{C'(x)x^2-2xC(x)}{x^4}+\frac{2C(x)}{x^3}=\frac{3}{x^2} \Rightarrow$$
$$C'(x)=3 \Rightarrow C(x)=3x+A$$
Thus: $y=\frac{C(x)}{x^2}=\frac{3x+A}{x^2}$ is a general solution.
To find a particular solution (that is, the value of the constant $A$), use the initial value $y(2)=-1:$
$$\frac{3\cdot 2+A}{2^2}=-1 \Rightarrow A=-10.$$
Thus: $y=\frac{3x-10}{x^2}$ is a particular solution.
