Question

$\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2} \\ \Rightarrow \frac{dy}{dx}={\left(\frac{y}{x}\right)}^2-\frac{y}{x} \\ \mathrm{Putting}y=vx,\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \therefore v+x\frac{dv}{dx}=v^2-v \\ \Rightarrow x\frac{dv}{dx}=v^2-2v \\ \Rightarrow \frac{1}{v^2-2v}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{v^2-2v}dv=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{v^2-2v+1-1}dv=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{{\left(v-1\right)}^2-{\left(1\right)}^2}dv=\int \frac{1}{x}dx \\ \Rightarrow \frac{1}{2}\log \left|\frac{v-1-1}{v-1+1}\right|\mathit{}=\log x+\log C \\ \Rightarrow \log \left|{\left(\frac{v-2}{v}\right)}^{\frac{1}{2}}\right|\mathit{}=\log Cx \\ \Rightarrow \log \left|{\left(\frac{{\displaystyle \frac{y}{x}-2}}{{\displaystyle \frac{y}{x}}}\right)}^{\frac{1}{2}}\right|\mathit{}=\log Cx \\ \Rightarrow \log \left|{\left(\frac{{\displaystyle y-2x}}{{\displaystyle y}}\right)}^{\frac{1}{2}}\right|\mathit{}=\log Cx \\ \Rightarrow {\left(\frac{{\displaystyle y-2x}}{{\displaystyle y}}\right)}^{\frac{1}{2}}=Cx \\ \Rightarrow \frac{{\displaystyle y-2x}}{{\displaystyle y}}\mathit{}=C^2{x}^2 \\ \Rightarrow y-2x=k{x}^{2}y,\mathrm{where}k=C^2 \end{gathered}$$