Integrate the differential equation (2y2 - x2) y′ + 3xy = 0

Integrate the following differential equation:

$(2y^2 - x^2) y' + 3xy = 0.$

As in the previous two exercise, we want to make the substitution $y = vx$ which gives us $y' = v + v'x$ . Therefore, we have

$\begin{align*} &&(2y^2 - x^2) y' + 3xy &= 0 \\[9pt] \implies && y' &= \frac{3xy}{x^2-2y^2} \\[9pt] \implies && y' &= \frac{3}{\frac{x}{y} - 2 \frac{y}{x}} \\[9pt] \implies && v + v'x &= \frac{3}{\frac{1}{v} - 2v}} \\[9pt] \implies && v'x &= 2 \left( \frac{v^3 + v}{1-2v^2} \right). \end{align*}$

Now, we separate variables and integrate,

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

Now, we substitute back $v = \frac{y}{x}$ ,

$\begin{align*} && \log (v^2) - 3 \log (v^2+1) &= \log (x^4) + C \\[9pt] \implies && \log \frac{y^2}{x^2} - 3 \log \left( \frac{y^2}{x^2} + 1 \right) &= \log (x^4) + C \\[9pt] \implies && \log y^2 - \log x^2 - 3 \log (x^2+y^2) + 3 \log x^2 &= \log x^4 + C \\[9pt] \implies && \log y^2 - \log x^2 - \log (x^2+y^2)^3 + \log x^6 &= \log x^4 + C \\[9pt] \implies && \log y^2 &= \log (x^2+y^2)^3 + C \\[9pt] \implies && y^2 &= C (x^2+y^2)^3. \end{align*}$

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Integrate the differential equation (2y² – x²) y′ + 3xy = 0

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