Plot `y=e^(x),y=e^(x)+1 and y=e^(x)-1.`

The equation of one of the curves whose slope of tangent at any point is equal to y+2x is (A) y=2(e^(x)+x-1) (B) y=2(e^(x)-x-1) (C) y=2(e^(x)-x+1) (D) y=2(e^(x)+x+1)

If e^(x)+e^(y)=e^(x+y)" then "y_(1)=

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

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

The solution of the differential equation dy/dx=e^(y-x)+e^(y+x) ; y(0) = 0 is (1) y=e^x(x+1) (2) e^-y=(e^-x-e^x)+1 (3) e^-y =(e^-x-e^x)-1 (4) e^-y =(e^-x+e^x)+1

The solution of the differential equation (dy)/(dx)+1=e^(x+y), is a. (x+y)e^(x+y)=0 b. (x+C)e^(x+y)=0 c. (x-C)e^(x+y)=1 d. (x+C)e^(x+y)+1=0

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 y = (e^(x)-e^(-x))/(e^(x)+e^(-x)) then prove that y = (e^(2x)-1)/(e^(2x)+1) .

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