First Order Separable differential Equation 
Problem:
  Solve the following differential equation:
  \begin{eqnarray*}
6x^2y \, dx - (x^3 + 1) \, dy &=& 0 \\
\end{eqnarray*}

Answer:
This is a separable differential equation.
\begin{eqnarray*}
\frac{6x^2}{x^3+1} \, dx - \frac{dy}{y} &=& 0 \\
\int \frac{6x^2}{x^3+1} \, dx - \int \frac{dy}{y} &=& c_1 \\
2 \ln{|x^3+1|} - \ln{|y|} &=& c_1 \\
\ln{(x^3+1)^2} - \ln{|y|} &=& c_1 \\
\ln{ \Big( \frac{(x^3+1)^2}{|y|} \Big) } &=& c_1 \\
(x^3+1)^2 &=& c|y| \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
(x^3+1)^2 &=& |cy| \\
\end{eqnarray*}
Is my answer different from the book's answer? I believe it is. What am I missing?Any idea of how to proceed?
Thanks,
Bob
 A: Both answers are correct.
Your answer $$(x^3+1)^2 = c |y|$$ makes the assumption that $c\ge 0$
The book's answer $$(x^3+1)^2 = |cy|$$ is OK for all values of $c$.
Thus to make sure that you can take any   value for c go with the book's answer, otherwise mention that $c\ge 0$ .  
A: $$\ln (x^3+1)^2=c_1+ \ln |y|$$
For $y  > 0$
$$\ln (x^3+1)^2=c_1+ \ln y$$
$$(x^3+1)^2=e^{c_1} y$$
For $y  <  0$
$$\ln (x^3+1)^2=c_1+ \ln (-y)$$
$$(x^3+1)^2=-e^{c_1} y$$
so we have that
$$(x^3+1)^2=|e^{c_1} y|=e^{c_1}|y|$$
On the other hand you have that
$$|ab| = |a||b|$$
so
$$|cy| = |c||y|$$
When in your book they choose $c=-2$ you take $e^{c_1}=2$ thats all
A: \begin{align}
-\infty &\lt c_1 &\lt \infty\\
0 &\lt e^{c_1} = c &\lt \infty\\
\text{Therefore, } c&= |c|
\end{align}
A: Solve the equation by separate $x,y$ :
$$\int\frac{6x^2}{x^3+1}dx=\int\frac{dy}{y}\\
\Rightarrow2\ln|x^3+1|+C_1=\ln|y|+C_2\\
\Rightarrow e^{2\ln|x^3+1|+C_1}=e^{\ln|y|+C_2}\\
\Rightarrow |x^3+1|^2=\frac{e^{C_2}}{e^{C_1}}|y|$$
Because $c=\frac{e^{C_2}}{e^{C_1}}>0$, $|cy|=c|y|$ if $c>0$ is specified. 
