设u = y/x则dy = udx + xdu则dy/dx = u + xdu/dx代回原式得u + xdu/dx = 1/2 + u/2 解得1/(1-u)^2 = 2x + C 则解为1/(1-y/x)^2 = 2x + C方程x^3 + ax + b = 0的解的情况:当a >= 0时有两个解.当 a < 0时由3x^2 + a = ...
求dy/dx=(x+y)/2x的原函数 和 x^3+ax+b=0的解的情况
设u = y/x则dy = udx + xdu则dy/dx = u + xdu/dx代回原式得u + xdu/dx = 1/2 + u/2 解得1/(1-u)^2 = 2x + C 则解为1/(1-y/x)^2 = 2x + C方程x^3 + ax + b = 0的解的情况:当a >= 0时有两个解.当 a < 0时由3x^2 + a = ...