Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+y=y^2 \\ \Rightarrow x\frac{dy}{dx}=y^2-y \\ \Rightarrow \frac{1}{y^2-y}dy=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y^2-y}dy=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{y\left(y-1\right)}dy=\int \frac{1}{x}dx\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}....\left(1\right) \\ \mathrm{Let}\frac{1}{y\left(y-1\right)}=\frac{A}{y}+\frac{B}{y-1} \\ \mathit{\Rightarrow }1=A\left(y-1\right)+B\left(y\right) \\ \mathrm{Putting}\mathit{}y=0,\mathrm{we}\mathrm{get} \\ 1=-A \\ \mathit{\Rightarrow }A=-1 \\ \mathrm{Putting}y=1,\mathrm{we}\mathrm{get} \\ 1=B \\ \therefore \frac{1}{y\left(y-1\right)}=\frac{-1}{y}+\frac{1}{y-1} \\ \Rightarrow \int \frac{1}{y\left(y-1\right)}dy\mathit{=}\int \frac{-1}{y}\mathit{}dy+\int \frac{1}{y\mathit{-}\mathit{1}}dy\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}.....\left(2\right)\mathit{} \\ \mathrm{From}\left(1\right)\&\left(2\right),\mathrm{we}\mathrm{get} \\ \int \frac{-1}{y}\mathit{}dy+\int \frac{1}{y-1}dy\mathit{}\mathit{=}\int \frac{1}{x}dx\mathit{} \\ \mathit{\Rightarrow }-\log \left|y\right|+\log \left|y-1\right|\mathit{}=\log \mathit{}\left|x\right|+\log \mathit{}\mathrm{C} \\ \mathit{\Rightarrow }\log \mathit{}\left|\frac{y-1}{y}\right|\mathit{-}\mathit{}\log \mathit{}\left|x\right|=\log \mathit{}\mathrm{C} \\ \mathit{\Rightarrow }\log \left|\frac{y-1}{xy}\right|=\log \mathrm{C} \\ \Rightarrow \frac{\mathrm{y}-1}{\mathrm{x}\mathrm{y}}=\mathrm{C} \\ \mathit{\Rightarrow }y-1=\mathrm{C}xy \\ \mathrm{Hence},y-1=\mathrm{C}xy\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$