Question

解微分方程dy/dx=√(2x+3y)

Answer 1

解:微分方程为dy/dx=√(2x+3y)，化为3dy/dx=3√(2x+3y)，有d(3y)/dx=3√(2x+3y)，设√(2x+3y)=u，微分方程化为d(u²-2x)/dx=2u，2udu/dx-2=2u，udu/dx=u+1，udu/(u+1)=dx，du-du/(u+1)=dx，u-ln|u+1|=x+ln|c|，u-x=ln[c(u+1)]，微分方程的通解为√(2x+3y)-x=ln[c(√(2x+3y)+1] (c为任意非零常数)

Answer 2

原函数只能得到隐函数表达式。

追问

dx=2/3(1-2/(3u+2))du？

追答

Answer 3

u = 2x+ 3y
du/dx = 2 + 3dy/dx
dy/dx=√(2x+3y)
(1/3)[du/dx -2 ] = √u
du/dx -2 = 3√u
∫du/(3√u+2) = ∫dx
x = ∫du/(3√u+2)
x= (2/3)√u - (4/3)ln|3√u+2| + C
=(2/3)√(2x+3y) - (4/3)ln|3√(2x+3y)+2| + C
let
v=√u
dv = du/(2√u)
du =2vdv
∫du/(3√u+2)
=∫2vdv/(3v+2)
= (2/3)∫ dv - (4/3)∫ dv/(3v+2)
= (2/3)v - (4/3)ln|3v+2| + C
= (2/3)√u - (4/3)ln|3√u+2| + C