求解微分方程：xy^'+y=x^2+3x+2

Question

如题所述

Answer 1

当 x = 0 时， y = 2；
当 x ≠ 0 时， (xy)' = x^2+3x+2
xy = ∫(x^2+3x+2)dx + C

= (1/3)x^3+(3/2)x^2+2x + C,
y = [ (1/3)x^3+(3/2)x^2+2x + C]/x

Answer 2

(xy)'=x^2+3x+2
xy=x^3/3+3x^2/2+2x+C
y=x^2/3+3x/2+2+C/x本回答被提问者采纳