Question

(1 − x²) dy + xy dx = xy² dx

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1-x^2\right)dy+xydx=x{y}^{2}dx \\ \Rightarrow \left(1-x^2\right)dy=x{y}^{2}dx-xydx \\ \Rightarrow \left(1-x^2\right)dy=xy\left(y-1\right)dx \\ \Rightarrow \frac{1}{y\left(y-1\right)}dy=\frac{x}{1-x^2}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y\left(y-1\right)}dy=\int \frac{x}{1-x^2}dx.....\left(1\right) \\ \mathrm{Considering}\mathrm{LHS}\mathrm{of}\left(1\right), \\ \mathrm{Let}\frac{1}{y\left(y-1\right)}=\frac{A}{y}+\frac{B}{y-1} \\ \Rightarrow 1=A\left(y-1\right)+By.....\left(2\right) \\ \mathrm{Substituting}y=1\mathrm{in}\left(2\right), \\ 1=B \\ \mathrm{Substituting}y=0\mathrm{in}\left(2\right), \\ 1=-A \\ \Rightarrow A=-1 \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}A\mathrm{and}B\mathrm{in}\frac{1}{y\left(y-1\right)}=\frac{A}{y}+\frac{B}{y-1},\mathrm{we}\mathrm{get} \\ \frac{1}{y\left(y-1\right)}=\frac{-1}{y}+\frac{1}{y-1} \\ \Rightarrow \int \frac{1}{y\left(y-1\right)}dy=\int \frac{-1}{y}dy+\int \frac{1}{y-1}dy \\ =-\log \left|y\right|+\log \left|y-1\right|+C_1 \\ \mathrm{Now},\mathrm{considering}\mathrm{RHS}\mathrm{of}\left(2\right),\mathrm{we}\mathrm{have} \\ \int \frac{x}{1-x^2}dx \\ \mathrm{Here},\mathrm{putting}1-x^2=t,\mathrm{we}\mathrm{get} \\ -2xdx=dt \\ \therefore \int \frac{x}{1-x^2}dx=\frac{-1}{2}\int \frac{1}{t}dt \\ =\frac{-1}{2}\log \left|t\right|+C_2 \\ =\frac{-1}{2}\log \left|1-x^2\right|+C_2\left[\because t=1-x^2\right] \\ \mathrm{Now},\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}\int \frac{1}{y\left(y-1\right)}dy\mathrm{and}\int \frac{x}{1-x^2}dx\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ -\log \left|y\right|+\log \left|y-1\right|+C_1=\frac{-1}{2}\log \left|1-x^2\right|+C_2 \\ \Rightarrow -\log \left|y\right|+\log \left|y-1\right|=-\frac{1}{2}\log \left|1-x^2\right|+C \\ \mathrm{where} \\ C=C_2-C_1 \end{gathered}$$