Question

Solve the Differential Equation, $(1+x)ydx+(1-y)xdy=0$.

Answer 1

Differential equation is $(1+x)ydx+(1-y)xdy=0$

$\Rightarrow (1+x)ydx=-(1-y)dx$

$\Rightarrow \left(\frac{1-x}{x}\right)dx=-\left(\frac{1-y}{y}\right)dy$

$\Rightarrow (\frac{1}{x}+1)dx=(\frac{-1}{y}+1)dy$.

On integrating both sides

$\therefore \int (\frac{1}{x}+1)dx=\int (\frac{-1}{y}+1)dy$

$\Rightarrow \int \frac{1}{x}dx+\int 1dx=-\int \frac{1}{y}dy+\int 1dy$

$\Rightarrow {\log }_{e}x+x=-{\log }_{e}y+y+c$

$\Rightarrow {\log }_{e}x+{\log }_{e}y=y-x+c$

$\Rightarrow {\log }_e(x.y)=y-x+c$

$\Rightarrow x.y=e^{y-x+c}$

This is required solution.