If `dy/dx=(e^y-x)^(-1),` where `y(0)=0` , then `y` is expressed explicitly as

(dy)/(dx) = e^(2x-y) + x^(3) e^(-y)

If x y=e^((x-y)), then find (dy)/(dx)

Find (dy)/(dx) where y=(tanx)/(x)

If x=e^(y+e^(y+e^y....oo) , where x > 0,then find (dy)/(dx)

If (2+sinx)(dy)/(dx)+(y+1)cosx=0 and y(0)=1 , then y((pi)/(2)) is equal to

If (dy)/(dx)-y log_(e) 2 = 2^(sin x)(cos x -1) log_(e) 2 , then y =

(dy)/(dx) = y tan x, y = 1 when x = 0

y(x^(2)y+e^(x))dx-e^(x)dy=0

Consider the differential equation (dy)/(dx)=e^(x-y+)x^(2)e^(-y) i . Express the equation in the form f(x)dx=g(y)dy. ii . Hence solve the equation.