# Solve the differential equation 2xlnx dy/dx + y = 0?

Sep 23, 2017

$y = \frac{A}{\sqrt{\ln x}}$

#### Explanation:

We have:

$2 x \ln x \frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$ ..... [A]

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{\mathrm{dy}}{\mathrm{dx}} + P \left(x\right) y = Q \left(x\right)$

So rewrite the equations in standard form as:

$\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{2 x \ln x} y = 0 \ldots . . \left[B\right]$

Then the integrating factor is given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus \frac{1}{2 x \ln x} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(\frac{1}{2} \ln | \ln x |\right)$ (see notes at end)
$\setminus \setminus = \exp \left(\ln \sqrt{| \ln x |}\right)$
 \ \ = sqrt(lnx))

And if we multiply the DE [B] by this Integrating Factor, $I$, we will have a perfect product differential form of $\left[A\right]$;

$\sqrt{\ln x} \frac{\mathrm{dy}}{\mathrm{dx}} + \sqrt{\ln x} \frac{1}{2 x \ln x} y = 0$
$\therefore \frac{d}{\mathrm{dx}} \left(y \sqrt{\ln x}\right) = 0$

 :. ysqrt(lnx)) = A

Which we can rearrange to get:

$\therefore y = \frac{A}{\sqrt{\ln x}}$

Which, is the General Solution.