How to find $x\frac{dy}{dx} = y+\frac{4}{x}$? How to find $x\frac{dy}{dx} = y+\frac{4}{x}$? Its given that $x>1, y=1$ when $x=1$
I dont think I can use separation of variables, for other methods, I don't really understand too well


*

*What does homologous of degree 0 mean? In my lecture notes, theres some mention of substitution of $z=...$. Resulting in an expression in $z$, which is then separable. 

*Integrating factor wasn't explained clearly either, or I just didn't really get it ... 


Which method can be used here and how? 
 A: You want to solve $xy'=y+\frac4x$, which is equivalent to
$$\begin{align}
xy'-y=\frac4x\\
\frac{xy'-y}{x^2}=\frac4{x^3}
\end{align}$$
Can you notice derivative of some familiar expression there? Are you able to continue from there?
Of course, there are many different methods, I have no doubt that you'll get many useful hints from other users, too.
A: Hint: it's a linear differential equation.
A: Integrating factors I can explain.  You may be familiar with the product rule for derivatives.
$d(uv)=udv+vdu$
The idea behind an integrating factor is to get one side of the equation to look like $(uy)'=uy'+u'y$, at which point you'll be able to integrate both sides.  First, let's get all y's on one side.
$xy'-y=\frac4x$
$y'-\frac yx=\frac4{x^2}$
Now we have 
$\frac{u'}u=-\frac1x$
$\ln u=-\ln x =\ln\frac1x$
$u=\frac1x$
If you multiply both sides by $\frac1x$, you'll get what Martin has in his answer.  You can verify that the left side is equal to $(\frac yx)'$.
A: You can also try it like this,
$$x\frac{dy}{dx} = y+\frac{4}{x}$$
$$xdy=ydx +\frac{4}{x}dx$$
$$xdy-ydx=\frac{4}{x}dx$$
dividing by $x^2$on both sides, we get,
$$\frac{xdy-ydx}{x^2}=\frac{4}{x^3}dx$$
now we know that  ,$d(\frac{y}{x})=\frac{xdy-ydx}{x^2}$
hence , upon integration we have,
$$\int d(\frac{y}{x})=\int \frac{4}{x^3}dx$$
$$\frac{y}{x}=\frac{-1}{2x^2} +c$$
$$y+\frac{1}{2x}=cx$$
A: As an alternative to the other answer, it is always worthwile to try and solve the homogenous problem $xz'=z/x$ first which has $z=x$ as a solution and then doing variation of parameters on $y:=v(x)z(x)$, which gives:
$x(v'x+v)=vx+4/x$
which should simplify nicely. Note that this takes advantage of the fact that your equation consists of a homogeneous part, $xy'-y$. That's why variation of parameters will always work. 
A: I would approach this as follows:
Rearrange the equation to have the following form:
$\displaystyle \frac{dy}{dx} + P(x)y = Q(x)$
$\therefore$ $\displaystyle \frac{dy}{dx} -\frac{y}{x} = \frac{4}{x^2}$
Then $\displaystyle P(x) = -\frac{1}{x}$ and $\displaystyle Q(x) = \frac{4}{x^2}$
Then assume an integrating factor has the form:
$\displaystyle \mu(x) = e^{\int P(x)dx}$
You can then solve the equation by using the following equation:
$\displaystyle y = \frac{1}{\mu(x)}\int\mu(x)Q(x)dx$
Your answer should be:
$\displaystyle y = -\frac{2}{x} + cx$
I'm afraid I can't expand on the assumption with the integrating factor, just gleaned from lecture notes (note to self, investigate later)
Happy to put the working if this is not helpful.
