求yy''=2(y'^2-y')满足初始条件y(0)=1,y'(0)=2的特解还有具体步骤

Question

求yy''=2(y'^2-y')满足初始条件y(0)=1,y'(0)=2的特解 还有具体步骤

Answer 1

设y' = p
那么y'' = dp/dx = dp/dy * dy/dx = pdp/dy
所以ypdp/dy = 2(p² - p)
化简得dy/y = dp/2(p-1)
积分得2lny = ln(p-1) +c
也就是y² = C(y'-1)
将y(0) = 1,y'(0) = 2代入得C = 1
所以dy/dx = y²+1
也就是dy/(y²+1) = dx
积分得acrtany = x+C
也就是y = tan(x+C)
将y(0) = 1代入得
C = π/4
所以y = tan(x+π/4)