求解一道微分的题目

2024-11-16 11:51:30
推荐回答(1个)
回答1:

这是典型的一类微分方程 dy/dx + P(x)y = Q(x)
因为 P(x) = -2/(x+1)
对P(x)积分得到 ∫P(x) dx = -2ln(x+1)
积分因子为:I(x) = e^∫P(x) dx = e^[-2ln(x+1)] = 1/(x+1)²
将原微分方程两侧同乘以积分因子,得到:
y'/(x+1)² - 2y/(x+1)³ = (x+1)^½
左侧变成全微分:
d[y/(x+1)²]/dx = (x+1)^½
d[y/(x+1)²] = (x+1)^½ dx
∫d[y/(x+1)²]/dx = ∫(x+1)^½ dx
y/(x+1)² = (2/3)(x+1)^(3/2) + c, c 为积分常数
y = (2/3)(x+1)^(7/2) + c(x+1)²