Question

(x² − y²) dx − 2xy dy = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(x^2-y^2\right)dx-2xydy=0 \\ \Rightarrow \frac{dy}{dx}=\frac{x^2-y^2}{2xy} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{x^2-{\left(vx\right)}^2}{2x\left(vx\right)} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{x^2-v^2{x}^2}{2v{x}^2} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{1-v^2}{2v} \\ \Rightarrow x\frac{dv}{dx}=\frac{1-v^2}{2v}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2}{2v} \\ \Rightarrow \frac{2v}{1-3v^2}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{2v}{1-3v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{3}\int \frac{-6v}{1-3v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{3}\log \left|1-3v^2\right|=\log \left|x\right|+\log C \\ \Rightarrow \log \left|1-3v^2\right|=-3\log \left|Cx\right| \\ \Rightarrow \log \left|1-3v^2\right|=\log \left|\frac{1}{{\left(Cx\right)}^3}\right| \\ \Rightarrow 1-3v^2=\frac{1}{{\left(Cx\right)}^3} \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ 1-3{\left(\frac{y}{x}\right)}^2=\frac{1}{{\left(Cx\right)}^3} \\ \Rightarrow \frac{x^2-3y^2}{x^2}=\frac{1}{C^3{x}^3} \\ \Rightarrow x\left(x^2-3y^2\right)=\frac{1}{C^3} \\ \Rightarrow x\left(x^2-3y^2\right)=K\left(\mathrm{where},K=\frac{1}{C^3}\right) \end{gathered}$$