How to find a specific curve if the initial value is not given? Question:
Let $y(x)$ be the solution of the differential equation 
$x\cdot ln(x)\dfrac{dy}{dx}+y=2x\cdot ln(x)$, $x\ge1$.
Find $y(e)$.
Answer: $y(e) = 2$
Problem:
So I understand that this can be converted into a simple linear differential equation and found that the solution is:
$y\cdot ln(x)=2(x\cdot ln(x) - x) + C$
This is a family of curves. However for solving the question, I need a specific curve out of all these.
What I don't understand is how how do I find that particular curve  as the initial value of the function is not given.
 A: I'd say that the answer $y(e) = 2$ is wrong if the question is stated like this.
You have already found the correct general solution of the first-order linear ODE, with one constant of integration $C \in \mathbb{R}$. From this you do indeed get $y(e) = C$ as William Elliot has pointed out. Therefore, you can obtain any value for $y(e)$, depending on the value of $C$.
The reason why the particular solution with $C=2$ is of interest is that it is the only solution of this ODE for which the (right-sided) limit as $x \searrow 1$ is finite.
For the general solution $y(x) = \frac{2x(\ln(x)-1)+C}{\ln(x)}$, $C \in \mathbb{R}$, we have
\begin{equation}
\lim_{x \searrow 1} y(x) = \left\{ \begin{array}{ll}
-\infty, C < 2\\
0, C = 2\\
\infty, C > 2
\end{array}
\right..
\end{equation}
Therefore, if we assume that the right-sided limit of $y(x)$ is finite as $x$ approaches $1$, then there remains only the particular solution with $C=2$, which satisfies $y(e) = 2$.
But this assumption needs to be added to the question, otherwise the answer is wrong.
