Question

求一阶线性微分方程

求一阶线性微分方程dy/dx=y/y-x的通解

Answer 1

解：∵微分方程为dy/dx=y/(y-x)，化为
dx/dy=(y-x)/y
∴有dx/dy=1-x/y，ydx/dy+x=y，
d(xy)/dy=y，xy=y²/2+c
(c为任意常数)，方程的通解为
x=y/2+c/y

Answer 2

let
u= (x^3+1)y
du/dx = (x^3+1) dy/dx + 3x^2. y
//
y' +3x^2.y/(x^3+1) = y^2.(x^3+1). sinx
(x^3+1)y' +3x^2.y = y^2.(x^3+1)^2. sinx
du/dx = u^2 .sinx
∫ du/u^2 = ∫ sinx dx
1/u = cosx +C
1/[(x^3+1)y] = cosx +C
y(0) =1
1= 1 +C
=> C=0
1/[(x^3+1)y] = cosx
y= 1/[cosx .(x^3+1)]