Sylvia Byrd

2022-07-07

Problem:
Solve the following differential equation:
$\begin{array}{rcl}6{x}^{2}y\phantom{\rule{thinmathspace}{0ex}}dx-\left({x}^{3}+1\right)\phantom{\rule{thinmathspace}{0ex}}dy& =& 0\end{array}$
This is a separable differential equation.
$\begin{array}{rcl}\frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\frac{dy}{y}& =& 0\\ \int \frac{6{x}^{2}}{{x}^{3}+1}\phantom{\rule{thinmathspace}{0ex}}dx-\int \frac{dy}{y}& =& {c}_{1}\\ 2\mathrm{ln}|{x}^{3}+1|-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}\left({x}^{3}+1{\right)}^{2}-\mathrm{ln}|y|& =& {c}_{1}\\ \mathrm{ln}\left(\frac{\left({x}^{3}+1{\right)}^{2}}{|y|}\right)& =& {c}_{1}\\ \left({x}^{3}+1{\right)}^{2}& =& c|y|\end{array}$
However, the book gets:
$\begin{array}{rcl}\left({x}^{3}+1{\right)}^{2}& =& |cy|\end{array}$
Is my answer different from the book's answer? I believe it is.

fugprurgeil

$\left({x}^{3}+1{\right)}^{2}=c|y|$
makes the assumption that $c\ge 0$
$\left({x}^{3}+1{\right)}^{2}=|cy|$
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 $c\ge 0$

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