Integrate the differential equation x (y + 4x) y′ + y (x + 4y) = 0

Integrate the following differential equation:

$x(y+4x)y' + y(x+4y) = 0.$

First, we rewrite the given equation a bit looking to make the substitution $v = \frac{y}{x}$ . We have

$\begin{align*} x(y+4x)y' + y(x+4y) &= 0 & \implies && y' &= \frac{-xy - 4xy^2}{xy + 4x^2} \\[9pt] && \implies && y' &= \frac{-\frac{y}{x} - 4 \left( \frac{y}{x} \right)^2}{\frac{y}{x} + 4}. \end{align*}$

Now, let $v = \frac{y}{x}$ , which gives us $y' = v + v'x$ and so we have

$\begin{align*} &&v + v'x &= \frac{-v-4v^2}{v+4} \\[9pt] \implies && v'x &= -5 \left( \frac{v^2+v}{v+4} \right) \\ \implies && \int \frac{1}{x} \, dx &= -\frac{1}{5} \int \frac{v+4}{v^2+v} \, dv \\[9pt] \implies && -\log |x^5| + C &= 4 \log |v| - 3 \log |v + 1| \\[9pt] \implies && -\log |x^5| + C &= \log (y^4) - \log (x^4) - \log |x+y|^3 + \log |x|^3 \\[9pt] \implies && C + \log (y^4) + \log (x^4) &= \log |x+y|^3 \\[9pt] \implies && C x^4 y^4 &= (x+y)^3. \end{align*}$

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Integrate the differential equation x (y + 4x) y′ + y (x + 4y) = 0

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