Everything was fine until you've made a silly mistake. It should be $$\large{v+x\frac{dv}{dx}=-\frac{v}{2+3v^2}}$$$$\large{x\frac{dv}{dx}=-\frac{3v+3v^3}{2+3v^2}}$$$$\large{\frac{2+3v^2}{3v+3v^3}}dv=-xdx$$
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Integrating factor is $\mu=y^3$. Then solution of equation $$xydx+(2x^2+3y^2)dy=0$$ is
$$y^6+x^2y^4=C$$
Related Questions in ORDINARY-DIFFERENTIAL-EQUATIONS
Everything was fine until you've made a silly mistake. It should be
$$\large{v+x\frac{dv}{dx}=-\frac{v}{2+3v^2}}$$ $$\large{x\frac{dv}{dx}=-\frac{3v+3v^3}{2+3v^2}}$$ $$\large{\frac{2+3v^2}{3v+3v^3}}dv=-xdx$$