Solve the initial value problem y′=4x(y2+1)y' = 4x(y^2 + 1)y′=4x(y2+1) with y(0)=1y(0) = 1y(0)=1. What is the value of yyy as x→π8x \to \sqrt{\frac{\pi}{8}}x→8π?
∞\infty∞
111
000
−1-1−1