# 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?

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You probably wanted to write homogeneous instead of homologous. BTW I am not sure why integration tag would be suitable here; I've added differential-equations. –  Martin Sleziak Apr 10 '12 at 6:14
substitute : $y=zx$ –  pedja Apr 10 '12 at 6:15

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.

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Hint: it's a linear differential equation.

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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 $(\frac yx)'$.

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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,

$$\int d(\frac{y}{x})=\int \frac{4}{x^3}dx$$

$$\frac{y}{x}=\frac{-1}{2x^2} +c$$

$$y+\frac{1}{2x}=cx$$

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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.

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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:

$\displaystyle y = \frac{1}{\mu(x)}\int\mu(x)Q(x)dx$

$\displaystyle y = -\frac{2}{x} + cx$

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.

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