finding a particular solution of $\frac{dy}{dx}=\frac{3y}{x}$ I need a help with the solving the differential equation explicitly.
$\frac{dy}{dx}=\frac{3y}{x}$, $y(1)=1$
and here is what I have so far:
$\frac{1}{y}dy=\frac{3}{x}dx$ (separate the variables)
$\int\frac{1}{y}\,dy=\int\frac{3}{x}\,dx$ (integrate)
$\ln\vert y\vert =3\ln\vert x\vert +C$
substitute the initial condition and get $C=0$
$\ln\vert y\vert =3\ln\vert x\vert$
$e^{\ln\vert y\vert}=e^{3\ln\vert x\vert}$
$\vert y\vert=\vert x\vert ^3$
$y=\pm\vert x\vert ^3$
$y=\vert x\vert ^3$ (choosing + judging by the initial condition)
and this is where I'm stuck. The answer key says it's $y=x^3$.
-EDIT-
now the next part of the question is asking what the domain of the particular solution is. The answer says it's $(0,\,\infty)$. I understand the initial condition of $x=1$ is in there, but I don't understand why I need to choose it, ignoring $(-\infty,\,0)$. If the problem stated the particular solution has no removable discontinuity, then I agree with their answer. But for now, I keep getting the domain of $x\neq 0$.
-EDIT2-
After more thorough research about this, I found that a solution to a differentiable equation is supposed to be "continuously differentiable function". Thus, only choosing a continuous interval that contains the initial condition. This is well-discussed in DE textbooks, but rarely in Calculus textbooks (The Chair of AP Calculus Development Committee agrees on this). For example, Stewart's Calculus textbooks don't even mention about the domain of solutions (granted, those textbooks are not designed to prep for AP exams). But I just hope that this is discussed by most instructors.
Source: http://ecademy.agnesscott.edu/~lriddle/apcalculus/apcentral-domain.pdf
 A: The differential equation
$$
\frac{dy}{dx}=\frac{3y}{x}
\tag{a}$$
is very very bad when $x=0$, since dividing by zero is not allowed.  Remember that?
So, when solving this DE, you get at most a solution defined on $(0,+\infty)$ or a solution defined on $(-\infty,0)$.  From your initial condition $y(1)=1$ we can get the solution on $(0,+\infty)$.
Both answers
$$
y = x^3,\qquad y = |x|^3
$$
are solutions of the DE on the domain $(-\infty,0)\cup (0,+\infty)$.  
In fact, you get even more solutions by taking $y=x^3$ on $(0,+\infty)$ and $y=Cx^3$ on $(-\infty,0)$, where $C$ is any constant.
All these solutions extend continuously to the domain $(-\infty,+\infty)$.  But since $\frac{0}{0}$ is undefined, none of these is a solution of (a) at the point $0$. Of course they are solutions of the related DE
$$
x\;\frac{dy}{dx}=3y
\tag{b}
$$
A: Note that $$y=\pm\vert x\vert ^3$$
is the same  as  $$y=\pm x^3$$
With the initial condition of $$y(1)=1$$
$$ y= -x^3$$ is not acceptable.
Thus $$y=x^3$$ which satisfies the differential equation and the initial condition. 
A: Here is a "non-integrative" way of finding a particular solution just by inspecting the equation.
$$y^{\prime}= 3\frac{y}{x}$$
An obvious candidate for a function with such a property is a power function $y=x^{\alpha}$ as we know 
$$y^{\prime} = \alpha x^{\alpha-1} = \alpha \frac{y}{x} \rightarrow \alpha = 3 \stackrel{y(1)=1}{\longrightarrow}y=x^3$$
