Question

(y² − 2xy) dx = (x² − 2xy) dy

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(y^2-2xy\right)dx=\left(x^2-2xy\right)dy \\ \Rightarrow \frac{dy}{dx}=\frac{y^2-2xy}{x^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{v^2{x}^2-2v{x}^2}{x^2-2v{x}^2} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{v^2-2v}{1-2v} \\ \Rightarrow x\frac{dv}{dx}=\frac{3v^2-3v}{1-2v} \\ \Rightarrow \frac{1-2v}{3v^2-3v}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1-2v}{3v^2-3v}dv=\int \frac{1}{x}dx \\ \Rightarrow -\int \frac{2v-1}{3v^2-3v}dv=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{3}\int \frac{6v-3}{3v^2-3v}dv=\int \frac{1}{x}dx \\ \mathrm{Putting}3v^2-3v=t \\ \Rightarrow \left(6v-3\right)dv=dt \\ \therefore -\frac{1}{3}\int \frac{1}{t}dt=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{3}\log \left|t\right|=\log \left|x\right|+\log C \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}t,\mathrm{we}\mathrm{get} \\ -\frac{1}{3}\log \left|3v^2-3v\right|=\log \left|x\right|+\log C \\ \Rightarrow -\frac{1}{3}\log \left|v^2-v\right|-\frac{1}{3}\log 3=\log \left|x\right|+\log C \\ \Rightarrow -\frac{1}{3}\log \left|v^2-v\right|=\log \left|x\right|+\log C-\frac{1}{3}\log 3 \\ \Rightarrow -\frac{1}{3}\log \left|v^2-v\right|=\log \left|x\right|+\log C_1\left(\mathrm{where},\log C_1=\log C-\frac{1}{3}\log 3\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}v,\mathrm{we}\mathrm{get} \\ -\frac{1}{3}\log \left|{\left(\frac{y}{x}\right)}^2-\left(\frac{y}{x}\right)\right|=\log \left|x\right|+\log C_1 \\ \Rightarrow -\frac{1}{3}\log \left|\frac{y^2}{x^2}-\frac{y}{x}\right|=\log \left|C_{1}x\right| \\ \Rightarrow \log \left|\frac{y^2-xy}{x^2}\right|=-3\log \left|C_{1}x\right| \\ \Rightarrow \log \left|\frac{y^2-xy}{x^2}\right|=\log \left|\frac{1}{{C_1}^3{x}^3}\right| \\ \Rightarrow \frac{y^2-xy}{x^2}=\frac{1}{{C_1}^3{x}^3} \\ \Rightarrow x{y}^2-x^{2}y=\frac{1}{{C_1}^3} \\ \Rightarrow x^{2}y-x{y}^2=-\frac{1}{{C_1}^3} \\ \Rightarrow x^{2}y-x{y}^2=K\left(\mathrm{where},\log K=-\frac{1}{{C_1}^3}\right) \end{gathered}$$