हल कीजिए `(e^x +1) y dy = ( y + 1)e^x dx`.

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

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 .

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

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

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

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

If y = (e^x - e^-x)/(e^x + e^-x) . Prove that dy/dx = 1 - y^2