Pam Stokes

2022-01-20

How to find $x\frac{dy}{dx}=y+\frac{4}{x}$?
It's given that .

macalpinee3

You want to solve $x{y}^{\prime }=y+\frac{4}{x}$, which is equivalent to
$x{y}^{\prime }-y=\frac{4}{x}$
$\frac{x{y}^{\prime }-y}{{x}^{2}}=\frac{4}{{x}^{3}}$
Can you notice derivative of some familiar expression there? Are you able to continue from there?

godsrvnt0706

Integrating factors I can explain. You may be familiar with the product rule for derivatives.
$d\left(uv\right)=udv+udu$
The idea behind an integrating factor is to get one side of the equation to look like ${\left(uy\right)}^{\prime }=u{y}^{\prime }+{u}^{\prime }y$, at which point you'll be able to integrate both sides. First, let's get all y's on one side.
$x{y}^{\prime }-y=\frac{4}{x}$
${y}^{\prime }-\frac{y}{x}=\frac{4}{{x}^{2}}$
Now we have
$\frac{{u}^{\prime }}{u}=-\frac{1}{x}$
$\mathrm{ln}u=-\mathrm{ln}x=\mathrm{ln}\frac{1}{x}$
$u=\frac{1}{x}$
If you multiply both sides by $\frac{1}{x}$, you'll get what Martin has in his answer. You can verify that the left side is equal to ${\left(\frac{y}{x}\right)}^{\prime }$.

RizerMix

As an alternative to the other answer, it is always worthwile to try and solve the homogenous problem $x{z}^{\prime }=\frac{z}{x}$ first which has $z=x$ as a solution and then doing variation of parameters on $y:=v\left(x\right)z\left(x\right)$, which gives:$x\left({v}^{\prime }x+v\right)=vx+\frac{4}{x}$which should simplify nicely. Note that this takes advantage of the fact that your equation consists of a homogeneous part, $x{y}^{\prime }-y$. That's why variation of parameters will always work.