Anonymous

dy/dx=1-y => dy/dx+y=1, this is a linear diff. equation with the integrating factor=e^x. Thus, e^xdy+ye^xdx=e^xdx => d(ye^x)=e^xdx => ye^x=e^x+C (after integration) => y=Ce^x+1 is the general solution.

Anonymous

∫dy/(1-y) = ∫dx -ln(1-y) = x+C y-1 = e^(x+C) = Ce^x y = 1+Ce^x

Anonymous

dy/dx = 1 - y Hence dy/(1-y) =dx Integrating both sides wrt 'x'gives -ln(1-y) =x +C ln(1-y)=-(x+C) Hence we get e^-(x+C)=1-y