(e^(x)+1)ydy=(y+1)e^(x)dx

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

(e^(x)+1)ydy-(y^(2)+1)e^(x)dx=0 , given y=0 when x=0

If y =( e^(x) + 1)/( e^(x)) ,then (dy)/(dx) =

The solution of the differential equation (1+e^(x))y(dy)/(dx) = e^(x) when y=1 and x = 0 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)

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

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 .

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

a^(x)(y^(2)+1)dx=ydy A) a^(x)=tan^(-1)y+c B) tan^(-1)y=a^(x)log_(a)e+c C) a^(x)log_(a)e=log(y^(2)+1)+c D) (a^(x))/(loga)=logsqrt(y^(2)+1)+c