Let y=f(x) satisfy the differential equation dydx=2xe-y, ∀x∈R. If y'(1)=1 then the number of solutions of the equation f(x)=f'(x) in (0,∞) is
(1)
∫ey dy=∫2x dx⇒f(x)=ln(x2+c)
f'(1)=1⇒C=1