求解微分方程
原方程 y" + y^2 = 1 设y'=p, 有y"=p*(dp/dy) 带入方程得 y’=p, y’’=p*(dp/dy) p*(dp/dy) + y2 = 1 p*dp = (1-y2) *dy p = sqrt(2*(y – y3/3 + C1)) dy/dx = sqrt(2*(y – y3/3 + C1)) dx = dy/sqrt(2*(y – y3/3 + C1)) x =∫dy/sqrt(2*(y – y3/3 + C1)) +C2 C1,C2为两个积分常量