1st order differential equation I am given the following:
$$
\begin{cases}
x \ln x \frac{dy}{dx}+y + x = 0, &\mbox{if}\quad  x>1, \\ 
y = 0, &\mbox{if} \quad x=e
\end{cases}
$$
I tried to separate it and got this:
$$
-y \ dy = \frac{-x}{x \ln x}dx
$$
Then i integrated it:
$$
\frac{-y}{2} = \frac{-1}{\ln x} \ + \ c, \\
{y} = 2 \left(\frac{1}{\ln x}+c \right)
$$
I tried solving after that, however, I am not getting the right answer, which is: 
$$
y=\frac{e-x}{\ln x}
$$
Could you help me identify the mistake?
Thank you very much for your help!!! :)
 A: Your mistake is in the separation. This isn't a separable differential equation - it is a first order linear equation. You can use an integrating factor to solve, or divide the entire differential equation by $x$ and you get:
$$(\ln x)y' + \frac{1}{x} y = -1$$
But then notice that this is equivalent to:
$$(y\ln x)' = -1$$
Integrate both sides and solve for $y$.
A: Rewrite:
$$
\begin{align}
x \ln x\ y'+y+x&=0\\
x \ln x\left(y'+\frac{y}{x \ln x}+\frac{1}{\ln x}\right)&=0\\
y'+\frac{y}{x \ln x}+\frac{1}{\ln x}&=0\\
y'+\frac{y}{x \ln x}&=-\frac{1}{\ln x}\tag1
\end{align}
$$
Equation $(1)$ is first-order linear ordinary differential equation. You have
$$
f(x)=\frac{1}{x \ln x}
$$
then
$$
\int f(x)\ dx=\int \frac{1}{x \ln x}\ dx=\int \frac{1}{\ln x}\ d(\ln x)=\ln(\ln x)+C.
$$
Hence, the integrating factor will be
$$
e^{\int f(x)\ dx}=e^{\ln(\ln x)+C}=e^C\ln x.
$$
Now, multiply $(1)$ by the integrating factor. It turns out to be
$$
\begin{align}
e^C\ln x\ y'+e^C\ln x\cdot\frac{y}{x \ln x}&=-e^C\ln x\cdot\frac{1}{\ln x}\\
\ln x\ y'+\frac{y}{x}&=-1\\
\frac{d}{dx}(\ln x\ y)&=-1\\
d(\ln x\ y)&=-dx\\
\int d(\ln x\ y)&=-\int\ dx\\
\ln x\ y&=-x+K\\
y&=\frac{K-x}{\ln x}.
\end{align}
$$
The last step, use the initial condition $y=0$ if $\ x=e$. You will obtain $K=e$. Thus
$$
\Large y=\Large\color{blue}{\frac{e-x}{\ln x}}.
$$
$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
A: your given D.E is linear in y:  
$$ \frac{dy}{dx}+\frac{y}{x \ln x} = -\frac{1}{x \ln x}$$,
so I.F: $e^{\int \frac{1}{x \ln x}dx} =lnx$
you get,on solving : y $ln$x=-x+c
from boundry conditions you get c=e
so the final answer becomes $y =\frac{e-x}{\ln x}$
