# How do you solve the differential y'xln(x)=y?

Nov 18, 2016

y=Ce^( 1/4x^2(2 ln x - 1) , where $C > 0$, and so is y.

#### Explanation:

To make ln x real, x > 0.

Separating variables and integrating,

$\int \frac{1}{y} \mathrm{dy} = \int x \ln x \mathrm{dx}$. So,

$\ln y$

$= \int \ln x d \left({x}^{2} / 2\right)$

$= \frac{1}{2} {x}^{2} \ln x - \frac{1}{2} \int {x}^{2} d \left(\ln x\right)$

$= \frac{1}{2} {x}^{2} \ln x - \frac{1}{2} \int {x}^{2} / x \mathrm{dx}$

$= \frac{1}{2} {x}^{2} \ln x - \frac{1}{2} \int {x}^{2} / x \mathrm{dx}$

$= \frac{1}{2} {x}^{2} \ln x - {x}^{2} / 4 + A$

So,

y =e^Ae^( 1/4x^2(2 ln x - 1)

=Ce^( 1/4x^2(2 ln x - 1) , where $C > 0$. and so is y.