Question

2xy dx + (x² + 2y²) dy = 0

Answer 1

$$\begin{gathered} 2xydx+\left(x^2+2y^2\right)dy=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{2xy}{x^2+2y^2} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=-\frac{2v{x}^2}{x^2+2v^2{x}^2} \\ \Rightarrow v+x\frac{dv}{dx}=-\frac{2v}{1+2v^2} \\ \Rightarrow x\frac{dv}{dx}=-\frac{2v}{1+2v^2}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{-3v-2v^3}{1+2v^2} \\ \Rightarrow \frac{1+2v^2}{3v+2v^3}dv=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1+2v^2}{3v+2v^3}dv=-\int \frac{1}{x}dx \\ \mathrm{Substituting}3v+2v^3=t,\mathrm{we}\mathrm{get} \\ 3\left(1+2v^2\right)dv=dt \\ \therefore \frac{1}{3}\int \frac{dt}{t}dv=-\int \frac{1}{x}dx \\ \Rightarrow \frac{1}{3}\log \left|t\right|=-\log \left|x\right|+\log \mathrm{C} \\ \Rightarrow \frac{1}{3}\log \left|3v+2v^3\right|=-\log \left|x\right|+\log \mathrm{C} \\ \Rightarrow \log \left|3v+2v^3\right|=-3\log \left|x\right|+3\log \mathrm{C} \\ \Rightarrow \log \left|\left(3v+2v^3\right)\times x^3\right|=\log {\mathrm{C}}^3 \\ \Rightarrow \left(3v+2v^3\right)\times x^3={\mathrm{C}}^3 \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \Rightarrow \left[\left(3\times \frac{y}{x}+2\times \frac{y^3}{x^3}\right)\times x^3\right]={\mathrm{C}}^3 \\ \Rightarrow 3y{x}^2+2y^3={\mathrm{C}}_1 \\ \mathrm{Hence},3y{x}^2+2y^3={\mathrm{C}}_1\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$