Question

(1 + x)ydx + (1+y)xdy = 0 solution? ​

Answer 1

Answer :

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

⇒(1+x)ydx=−(1−y)dx

⇒(1-x/x)dx = - (1-y/y)dy

⇒(1/x + 1)dx = (-1/y + 1)

By integrating both sides

∫(1/x + 1)dx = ∫(-1/y + 1)dy

∫1/x × dx + ∫1dx = -∫ 1/y × dy + ∫1dy

⇒log(e) x+x = −log(e) y+y+c

⇒log(e) x+log(e)y = y−x+c

⇒log(e) (x.y)=y−x+c

⇒x.y = e^y-x+c

proved

Answer 2

Answer:

no idea

Step-by-step explanation: