Q- 73. The general solution of the differential equation (ex + 1) ydy = (y + 1) ex dx is:
(A) (y + 1) = k (ex + 1) (B) y + 1 = ex + 1 + k
(C) y = log {k (y + 1) (ex + 1)} (D) y=logex+1y+1+k
Ans:(C)
Explanation:
Given differential equation (ex+1) ydy = (y+1) ex dx⇒dydx=ex1+yex+1y ⇒dxdy=ex+1yex1+y⇒dxdy=exyex1+y+yex1+y⇒dxdy=y1+y+y1+yex⇒dxdy=y1+y1+1ex⇒dxdy=y1+yex+1ex⇒y1+ydy=exex+1dx⇒On integrating both sides, we get⇒∫y1+ydy=∫exex+1dx⇒∫1+y-11+ydy=∫exex+1dx⇒∫dy-∫11+ydy⇒=∫exex+1dx⇒y-log1+y=logex+1+log k⇒y=log1+y+log1+ex+log k⇒y=logk1+y1+ex