What is the general solution of the differential equation y′=y2−y−2y' = y^2 - y - 2y′=y2−y−2?
y(t)=2y(t) = 2y(t)=2 and y(t)=−1y(t) = -1y(t)=−1 are the only solutions.
y(t)=2+3Ce3t−1y(t) = 2 + \frac{3}{Ce^{3t}-1}y(t)=2+Ce3t−13
y(t)=1+Cety(t) = 1 + Ce^ty(t)=1+Cet
y(t)=−1+Ce2ty(t) = -1 + Ce^{2t}y(t)=−1+Ce2t