First Order Separable differential Equation

Question

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

Answer 1

$$\ln (x^3+1)^2=c_1+ \ln |y|$$ 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

Answer 2

\begin{align} -\infty &\lt c_1 &\lt \infty\\ 0 &\lt e^{c_1} = c &\lt \infty\\ \text{Therefore, } c&= |c| \end{align}

Answer 3

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 $c\ge 0$ .

Answer 4

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.