Problem:

Solve the following differential equation:

$\begin{array}{rcl}6{x}^{2}y\phantom{\rule{thinmathspace}{0ex}}dx-({x}^{3}+1)\phantom{\rule{thinmathspace}{0ex}}dy& =& 0\end{array}$

Answer:

This is a separable differential equation.

$\begin{array}{rcl}\frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\frac{dy}{y}& =& 0\\ \int \frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\int \frac{dy}{y}& =& {c}_{1}\\ 2\mathrm{ln}|{x}^{3}+1|-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}({x}^{3}+1{)}^{2}-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}{\textstyle (}\frac{({x}^{3}+1{)}^{2}}{|y|}{\textstyle )}& =& {c}_{1}\\ ({x}^{3}+1{)}^{2}& =& c|y|\end{array}$

However, the book gets:

$\begin{array}{rcl}({x}^{3}+1{)}^{2}& =& |cy|\end{array}$

Is my answer different from the book's answer? I believe it is.

Solve the following differential equation:

$\begin{array}{rcl}6{x}^{2}y\phantom{\rule{thinmathspace}{0ex}}dx-({x}^{3}+1)\phantom{\rule{thinmathspace}{0ex}}dy& =& 0\end{array}$

Answer:

This is a separable differential equation.

$\begin{array}{rcl}\frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\frac{dy}{y}& =& 0\\ \int \frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\int \frac{dy}{y}& =& {c}_{1}\\ 2\mathrm{ln}|{x}^{3}+1|-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}({x}^{3}+1{)}^{2}-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}{\textstyle (}\frac{({x}^{3}+1{)}^{2}}{|y|}{\textstyle )}& =& {c}_{1}\\ ({x}^{3}+1{)}^{2}& =& c|y|\end{array}$

However, the book gets:

$\begin{array}{rcl}({x}^{3}+1{)}^{2}& =& |cy|\end{array}$

Is my answer different from the book's answer? I believe it is.