Question

微分方程通解？

yy''+y'²=1通解怎么求，请给出具体过程

Answer 1

这属于缺 x （自变量）型
令 y' = p, 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
则 ypdp/dy = 1-p^2, -2pdp/(1-p^2) = -2dy/y
ln(1-p^2) = -2lny + lnC1
1-p^2 = C1/y^2, p = dy/dx = ±√(1-C1/y^2)
dy/√(1-C1/y^2) = ±dx, ydy/√(y^2-C1) = ±dx
√(y^2-C1) = ±x + C2, y^2-C1 = (±x + C2)^2

Answer 2

Answer 3

追问

十分抱歉，你的答案不对

Answer 4