Solve : e^x (x+1)dx + (ye^y − xe^x )dy=0


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Question Number 41378 by rahul 19 last updated on 06/Aug/18

$Solve :$ $e^{x} (x + 1) d x + ({ye}^{y} - {x e}^{x}) dy = 0$
Commented by rahul 19 last updated on 06/Aug/18
If I put xe^x = t. does this helps?
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18

	$u = e^{x} x d u = e^{x} (x + 1) d x$ $d u + ({y e}^{y} - u) d y = 0$ $d u = (u - {y e}^{y}) d y$ $\frac{d u}{d y} - u = - {y e}^{y}$ $i n t r e g a t i n g f a c t o r e^{\int - d y} = e^{- y}$ $e^{- y} \frac{d u}{d y} + (- {u e}^{- y}) = - y$ $\frac{d (u . e^{- y})}{d y} = - y$ $d (u . e^{- y}) = - y d y$ $i n t r e g a t i n g$ $u . e^{- y} = - \frac{y^{2}}{2} + c$ $(e^{x} . x) . e^{- y} = - \frac{y^{2}}{2} + c$
	Commented by rahul 19 last updated on 07/Aug/18
	Thank You sir ��
	Commented by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18
	$checking... differentiating wrt to x e^(−y) {e^x (x+1)}+e^x .x(−e^(−y) )(dy/dx)=−y(dy/dx) multiply both side by e^y e^x (x+1)−e^x x(dy/dx)=−ye^y (dy/dx) e^x (x+1)dx+(e^y y−e^x x)dy=0$
	$c h e c k i n g . . .$ $d i f f e r e n t i a t i n g w r t t o x$ $e^{- y} {e^{x} (x + 1)} + e^{x} . x (- e^{- y}) \frac{d y}{d x} = - y \frac{d y}{d x}$ $m u l t i p l y b o t h s i d e b y e^{y}$ $e^{x} (x + 1) - e^{x} x \frac{d y}{d x} = - {y e}^{y} \frac{d y}{d x}$ $e^{x} (x + 1) d x + (e^{y} y - e^{x} x) d y = 0$
	Commented by rahul 19 last updated on 07/Aug/18
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