`(1+e^(x))/(y)(dy)/(dx)=e^(x),` when `y=1, x=0`

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

(x-y)(1-(dy)/(dx))=e^(x)

(dy)/(dx) -y =e^(x ) " when" x=0 and y=1

Solve : (1+e^(2x))dy+(1+y^(2))e^(x)dx=0 when y(0)=1

Solve the differential equation: e^(x//y)(1-(x)/(y))+(1+e^(x//y)) (dx)/(dy)=0 when x=0, y=1

Solve the following differential equation (5) (x+1)dy/dx-1=2e^(-y),y=0, when x=1,y=0.

(x+1)(dy)/(dx) -1 = 2e^(-y) , y=0, " when " x=1

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 .

Solution of e^((dy)/(dx)) = x when x = 1 and y = 0 is