Differential equation $\frac{dy}{dx}+\frac{2y}{x}=\frac{4}{x^2}$ (need help) 
Find the general solution of the differential equation:
$$\frac{dy}{dx}+\frac{2y}{x}=\frac{4}{x^2}$$


$$\frac{dy}{dx}x^2+2xy=4\tag1$$
$$x^2+\int 2xydx=\int 4dx\tag2$$
$$x^2+yx^2=4x+c\tag3$$
$$y=\frac{4x+c}{x^2+1}\tag4$$
The answer from the book is $$y=\frac{4}{x}+\frac{c}{x^2}$$
 A: $$\frac{dy}{dx}+\frac{2y}{x}=\frac{4}{x^2}\qquad . . . . . . (1)$$
This is a first order linear differential equation. 
Integrating factor (I.F.) is $$e^{\int \frac{2}{x} dx}=e^{2\log x}=x^2$$
So multiplying both side of $(1)$ by I.F. 
$$x^2\frac{dy}{dx}+2xy=4\implies \frac{d}{dx}(x^2y)=4$$
and then integrating$$x^2y=4x+c\implies y=\frac{4}{x}+\frac{c}{x^2}\qquad \text{where $c$ is integrating constant.}$$ 
A: You can solve firstly the homogeneous differential equation.
$\frac{dy}{dx}=-\frac{2y}{x}$
$\frac{dy}{y}=-\frac{2}{x}dx$
$\int \frac{dy}{y}=-\int\frac{2}{x}dx$
$\ln(y)=-2\cdot \ln(x)+c$
$\ln(y)=- \ln(x^2)+c$
$\ln(y)=\ln\left(\frac{1}{x^2}\right)+c$
$y_H=C\cdot \frac{1}{x^2}$
Variation of the constant.
$y_I=C(x)\cdot \frac{1}{x^2}$
Differentiating (product rule)
$y_I^{'}=C^{'}(x)\cdot \frac{1}{x^2}-C^{}(x)\cdot \frac{2}{x^3}$
Plugging into the differential equation
$C^{'}(x)\cdot \frac{1}{x^2}\underbrace{-C^{}(x)\cdot \frac{2}{x^3}+2\cdot C(x)\cdot \frac{1}{x^3}}_{=0}=\frac{4}{x^2}$
$C^{'}(x)\cdot \frac{1}{x^2}=\frac{4}{x^2}$
$C^{'}(x)=4\Rightarrow C(x)=4x$
Thus the solution is $y=y_H+y_I=C\cdot \frac{1}{x^2}+4x\cdot  \frac{1}{x^2}$
$$y=\frac{C}{x^2}+ \frac{4}{x}$$
