Solve the initial value problem y′=4x3y2y' = 4x^3 y^2y′=4x3y2 with y(0)=−1y(0) = -1y(0)=−1. What is the value of y(1)y(1)y(1)?
y(1)=−0.5y(1) = -0.5y(1)=−0.5
y(1)=−2y(1) = -2y(1)=−2
y(1)=0.5y(1) = 0.5y(1)=0.5
y(1)=−1y(1) = -1y(1)=−1