求微分方程y^2*y''+1=0积分曲线,使该积分曲线过点(0,1/2),且该点的切线斜率为2

问题描述:

求微分方程y^2*y''+1=0积分曲线,使该积分曲线过点(0,1/2),且该点的切线斜率为2

令 y ' = p = dy / dx,则 y'' = dp / dx = dp/dy * dy/dx = dp/dy * p,代入原方程得到:
y^2 * dp/dy * p + 1 = 0 => -pdp = 1/y^2 dy
两边同时积分上式得到:
-1/2 p^2 = -1/y + C1 => p^2 = 2/y + C1 (仍用C1记常数)
该积分曲线过点(0,1/2),且该点的切线斜率为2,代入上式有:
2^2 = 4 + C1 => C1 = 0 => p = dy/dx = √(2/y) => √y dy = √2 dx
积分上式化简即有:
x = √2/3 * y^(3/2) + C2
过点(0,1/2),代入得到:C2 = -1/6,所以积分曲线:
x = √2/3 * y^(3/2) - 1/6