Linear differential equation - inseparable? I need help with solving following linear differential equation:
$$y^\prime + \dfrac{3y}{x} = \dfrac{3}{x^4}$$

I have set the right side equal to $0$ and started solving:
$$y^\prime + \dfrac{3y}{x} = 0$$
$$y^\prime = - \dfrac{3y}{x}$$
$$\dfrac{dy}{dx}\times \dfrac{1}{y} = - \dfrac{3}{x}$$
$$\int\dfrac{1}{y} dy = -3\int \dfrac{1}{x} dx$$
$$\ln(y)=-3\times \ln(x) + c$$
And finally solution with left side (lets say this $c$ is: $e^c$)
$$y = c\times(-3x)$$

Using some formulas
$$y = c\times f(x)$$
$$y = c(x)\times f(x)$$
$$y^\prime = c^\prime(x)\times f(x)+c(x)\times f^\prime (x)$$
$$y = c(x)\times (-3x)$$
$$y^\prime = c^\prime(x)\times (-3x)+c(x)\times (-3)$$
And I replaced $y$ and $y^\prime $ in original equation and get
$$c^\prime(x)\times (-3x)+c(x)\times (-3)+\dfrac{3\times (c(x)\times(-3x))}{x} = \dfrac{3}{x^4}$$
And after adjustment:
$$\dfrac{c^\prime(x)\times (-3x)+c(x)\times (-3)+3\times (c(x)\times(-3x))}{x} = \dfrac{3}{x^5}$$
And what are next steps? I know I need to get $c(x)$ but how will I obtain it and what will be the final result?
 A: I would personally use the integrating factor method as @RobertZ has done, but here is an answer corresponding to the method you were planning to use:

You've made a mistake with the homogeneous solution. Going from here:
$$\ln(y)=-3\ln(x) + c\iff \ln(y)=\ln(x^{-3})+c$$
Hence, the homogeneous solution to your ODE will be:
$$e^{\ln{y}}=e^{\ln(x^{-3})+c} \iff y=e^c\cdot \ln(x^{-3})\iff y(x)=\frac{C}{x^3}$$
Where $C=e^c$.

It seems like the method you want to use is Variation of Parameters. To do this, you should assume $C$ is a function of $x$.
$$y(x)=\frac{C(x)}{x^3} \tag{1}$$
We then differentiate $(1)$ using the product rule:
$$y'(x)=\frac{C'(x)}{x^3}-\frac{3C(x)}{x^{4}} \tag{2}$$
Now, substitute $(1)$ and $(2)$ into your ODE:
$$y'+\frac{3y}{x}=\frac{3}{x^4} \implies \frac{C'(x)}{x^3}-\frac{3C(x)}{x^4}+\frac{3C(x)}{x^4}=\frac{3}{x^4}$$
The $C(x)$ terms should cancel, and you should be able to evaluate $C(x)$ by integrating both sides with respect to $x$.
$$\frac{C'(x)}{x^3}=\frac{3}{x^4}$$
$$\int C'(x)~dx=\int \frac{3}{x}~dx$$
After evaluating $C(x)$, just substitute it into $(1)$, and you will have the general solution.
A: By multiplying both sides by the integrating factor $x^3$, we get 
$$D(x^3y(x))=x^3y'(x) + 3 x^2 y(x) = \frac{3}{x}.$$
Hence
$$x^3y(x)=\int \frac{3}{x}\, dx=3\ln|x|+C,$$
that is
$$y(x)=\frac{3\ln|x|+C}{x^3}.$$
