" -3."(e^(x)+1)ydy+(y+1)dx=0

The solution of (e^(x) + 1) y dy + (y+ 1) dx = 0 is

If the curve satisfy the equation (e^(x)+1)ydy=(y+1)e^(x)dx, passes through (0,0) and (k,1) then k is

The general solution of differential equation (e^(x)+1)ydy=(y+1)e^(x)dx is

The solution of the equation (e^x+1)ydy+(y+1)dx=0 , is (A) e^(x+y)=C(y+1)e^x (B) e^(x+y)=C(x+1)(y+1) (C) e^(x+y)=C(y+1)(1+e^x) (D) e^(xy)=C(x+y)(e^x+1)

(e ^ (x) +1) is + (y + 1) dx = 0

The general solution of differential equation (e^(x)+1)ydy=(y+1)(e^(x))dx"is"

Solve the following differential equations: (e^x+1)ydy=(y+1)e^xdx

If e^(x) + e^(y) = e^(x + y) , then prove that (dy)/(dx) = (e^(x)(e^(y) - 1))/(e^(y)(e^(x) - 1)) or (dy)/(dx) + e^(y - x) = 0 .