Question

求微分方程x*dy/dx+y-3=0满足初始条件x=1,y=0的特解.急!大一高数!

Answer 1

x*dy/dx+y-3 = 0
x ≠ 0 时， 微分方程化为 dy/dx + y/x = 3/x 为一阶线性微分方程， 通解是
y = e^(-∫dx/x)[∫(3/x)e^(∫dx/x)dx + C]
= (1/x)(∫3dx + C) = 3 + C/x, x = 1, y = 0 代入得 C = -3
特解是 y = 3 - 3/x

Answer 2

公式:dy/dx+p(x)y=q(x)
通解公式:y=[∫q(x)e∧∫p(x)+c ]*e∧－∫p(x )
原式可变型为:dy/dx+y*1/x=3/x
p(x )=1/x q(x )=3/x