Question

高等数学微分方程求通解，速答

Answer 1

先求齐次方程dy/dx=y/x
dy/y=dx/x
ln|y|=ln|x|+ln|C|
即y=Cx
由常数变易法，令y=C(x)x
则y'=C'(x)x+C(x)
带入原方程dy/dx=y/x+x²得
C'(x)=x
则C(x)=x²/2+C
故原方程的通解为y=x(x²/2+C)
即y=x^3/2+Cx

Answer 2

let

u= y/x
du/dx = (1/x).dy/dx - y/x^2
dy/dx =x.[ du/dx + (1/x)u]
/
dy/dx -y/x =x^2
x.[ du/dx + (1/x)u] - u = x^2
x.du/dx =x^2
∫du = ∫xdx
u = (1/2)x^2 + C
y/x =(1/2)x^2 + C
y=(1/2)x^3 + Cx