ydx-xdy-1-x-2-y-2-xdx-0-

Question Number 100184 by bobhans last updated on 25/Jun/20

\frac{ydx + xdy}{1 - x^{2} y^{2}} + xdx = 0

Answered by maths mind last updated on 26/Jun/20

u = x y \Rightarrow d u = y d x + x d y

\Leftrightarrow \frac{d u}{1 - u^{2}} + x d x = 0

\Rightarrow \frac{l n (\frac{1 + u}{1 - u})}{2} = - \frac{x^{2}}{2} + c

\Rightarrow \frac{1 + u}{1 - u} = {k e}^{- x^{2}}

\Rightarrow u = \frac{{k e}^{- x^{2}} - 1}{1 + {k e}^{- x^{2}}}

y = \frac{1}{x} (\frac{{k e}^{- x^{2}} - 1}{1 + {k e}^{- x^{2}}})