<s>[INST] 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
 [/INST] $$\ln (x^3+1)^2=c_1+ \ln |y|$$
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
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