Question

一道求微分方程通解的题

一道求微分方程通解的题如图谢谢大佬

Answer 1

常数微分方程求解

Answer 2

如图

Answer 3

(x+1)y' -2y = (x+1)^(7/2)
let
u = y/(x+1)^2
(x+1)^2.u =y
(x+1)^2.du/dx + 2(x+1)u = dy/dx
------------

(x+1)y' -2y = (x+1)^(7/2)
y' - 2y/(x+1) = (x+1)^(5/2)
(x+1)^2.du/dx + 2(x+1)u - 2(x+1)u = (x+1)^(5/2)
(x+1)^2.du/dx =(x+1)^(5/2)
∫ du = ∫ (x+1)^(1/2) dx
u = (2/3) (x+1)^(3/2) + C
y/(x+1)^2 = (2/3) (x+1)^(3/2) + C
y= (2/3) (x+1)^(7/2) + C(x+1)^2