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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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    substitute : $y=zx$2012-04-10

6 Answers 6

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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:

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

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.