求(x+1)y'-y＝x的通解

Question

如题所述

Answer 1

(x+1)y'-y ＝ x
x = -1 时， y = 1；
x ≠ -1 时，y' - y/(x+1) ＝ x/(x+1) 为一阶线性微分方程
y = e^[∫dx/(x+1)]{∫[x/(x+1)]e^[-∫dx/(x+1)]dx + C}
= (x+1){∫[xdx/(x+1)^2] + C}
= (x+1){∫[1/(x+1) - 1/(x+1)^2]dx + C}
= (x+1)[ln|x+1| + 1/(x+1) + C]
= (x+1)ln|x+1|+ C(x+1) + 1]

Answer 2

p(x) = -1/(x+1)
∫p(x) dx = -ln|x+1| +C
e^[∫p(x) dx] = 1/(x+1)
(x+1)y'-y＝x
y'-[1/(x+1)]y＝x/(x+1)
两边乘以 1/(x+1)
[1/(x+1)].[y'-[1/(x+1)]y]＝x/(x+1)^2
d/dx [ y/(x+1)] = x/(x+1)^2
y/(x+1)
=∫ x/(x+1)^2 dx
=∫ [(x+1)-1 ]/(x+1)^2 dx
=∫ [1/(x+1) - 1/(x+1)^2 ] dx
= ln|x+1| +1/(x+1) + C
y=(x+1).ln|x+1| +1 + C.(x+1)