Integrate the following differential equation:
![\[ y' = \frac{y (x^2+xy+y^2)}{x(x^2 + 3xy + y^2)}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4fef13c9091090b188b232a0fbec6833_l3.png "Rendered by QuickLaTeX.com")
First, we rearrange the terms on the right hand side with the goal of making the substitution 
,
![\begin{align*} \frac{y(x^2+xy+y^2)}{x(x^2+3xy+y^2)} &= \frac{x^2y + xy^2 + y^3}{x^3 + 3x^2 y + xy^2} \\[9pt] &= \frac{ \left( \frac{y}{x} \right) + \left( \frac{y}{x} \right)^2 + \left( \frac{y}{x} \right)^3}{1 + 3 \left( \frac{y}{x} \right) + \left( \frac{y}{x} \right)^2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-08f16bfdc7eb88b931122d279f38d920_l3.png "Rendered by QuickLaTeX.com")
Now we make the substitution which gives us 
and so we have
![\begin{align*} && y' &= \frac{y(x^2+xy+y^2)}{x(x^2 + 3xy + y^2)} \\[9pt] \implies && v + v'x &= \frac{v + v^2 + v^3}{1 + 3v + v^2} \\[9pt] \implies && v'x &= \frac{v+v^2 + v^3 - v - 3v^2 - v^3}{1+3v + v^2} \\[9pt] \implies && v'x &= -\frac{2v^2}{1+3v+v^2} \\[9pt] \implies && \int \frac{1}{x} \, dx &= -\frac{1}{2} \int \frac{1+3v+v^2}{v^2} \, dv \\[9pt] \implies && \log |x| + C &= -\frac{1}{2} \left( -\frac{1}{v} + \log |v|^3 + v \right) \\[9pt] \implies && \log (x^2) + C &= \frac{x}{y} - \log \left(\frac{y^3}{x^3} \right) - \frac{y}{x} \\[9pt] \implies && C &= \frac{x}{y} - \frac{y}{x} + \log \left( \frac{x^3}{y^3} \right) + \log \left( \frac{1}{x^2} \right) \\[9pt] \implies && C &= \frac{x}{y} - \frac{y}{x} + \log \left( \frac{x}{y^3} \right) \\[9pt] \implies && C &= \frac{y}{x} - \frac{x}{y} + \log \left( \frac{y^3}{x} \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1796b738965d2a9597584f92a99898d0_l3.png "Rendered by QuickLaTeX.com")
Integrate the following differential equation:
First, we rearrange the terms on the right hand side with the goal of making the substitution ,
Now we make the substitution which gives us and so we have