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?

 A: Something is wrong in your procedure. In particular 
$$\frac{xy'}{2}+y\not=\frac{(xy)'}{2}.$$
Note that by multiplying both sides of the ODE by $x^2$, we get
$$x^2 y'+2xy=3$$
and the LHS is the derivative of $x^2 y$. Hence, after integrating,  we obtain the general solution:
$$y(x)=\frac{3x+C}{x^2}.$$ 
Finally, in order to find the constant $C$, use the condition $y(2)=-1$.
A: 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.
A: I have an itch to add a proper procedure.
Suppose we have the following first order linear non-homogeneous differential equation!
$$\frac{dy}{dx}+\frac{2y}{x}=\frac{3}{x^2}$$
By method of inspection we notice that we can use the integrating factor method since this is a linear one!
So,
Generally,
$$\frac{dy}{dx}+P(x)y=Q(x)$$
The integrating factor can be found by
$$\mu=e^{\int P(x)dx}$$
So, onto your question
Integrating factor is 
$$\mu=e^{\int \frac{2}{x}dx}$$
$$\mu=e^{2ln|x|}$$
$$\mu=e^{ln|x^2|}$$
$$\mu=x^2$$
Multiplying the differential equation with the integrating factor we have,
$$\frac{x^2dy}{dx}+2xy=3$$
So, by it we have 
$$\int d(yx^2)=\int 3 dx$$
$$yx^2=3x+C$$
Initial value is $y(2)=-1$
So by it we have
$$C=(-1)(2)^2-3(2)$$
$$C=-10$$
Rewrite the particular solution differential equation
$$yx^2=3x-10$$
Addendum. Some old story for integrating factor!
$$M(x,y)dx+N(x,y)dy=0$$
Know that if f is the differential equation!
$$M=\frac{\partial f}{\partial x},N=\frac{\partial f}{\partial y}$$
The differential equation is said to be exact if
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
It is also known as Clairaut's theorem. Which implies continuity of certain function!
Let it be the case where,
$$\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}$$
We multiply the differential equation with integrating factor $\mu(x)$
We assume that $\mu$ depends on one variable only not the other. So, in our case assume it to be dependent on x!
$$\mu(x)Mdy+\mu(x)Ndx=0$$
Applying Clairaut's theorem and assume after some multiplication with integrating factor we have an exact differential equation.
$$\mu$$
$$\frac{\partial \mu(x)(M(x,y))}{\partial y}=\frac{\partial \mu(x)(N(x,y))}{\partial x}$$
Since $\mu(x)$ is not affected by y then we can write,
$$\mu(x)\frac{\partial (M(x,y))}{\partial y}=\mu(x)\frac{\partial (N(x,y))}{\partial x}+N(x,y)\frac{\partial \mu}{\partial x}$$
$$\int\frac{1}{N(x,y)}(\frac{\partial (M(x,y))}{\partial y}-\frac{\partial (N(x,y))}{\partial x})dx=\int\frac{d\mu}{ \mu(x)}$$
$$e^{\int\frac{1}{N(x,y)}(\frac{\partial (M(x,y))}{\partial y}-\frac{\partial (N(x,y))}{\partial x})dx}=\mu$$
It is considered lucky if we can find the integrating factor of the differential equation!
