Solve y′=3y2y' = 3y^2y′=3y2 with y(0)=1y(0) = 1y(0)=1.
y=1/(1−3x)y = 1/(1-3x)y=1/(1−3x)
y=1/(1+3x)y = 1/(1+3x)y=1/(1+3x)
y=−1/(1−3x)y = -1/(1-3x)y=−1/(1−3x)
y=1−3xy = 1-3xy=1−3x