Question

(y^2-2xy)dx+x^2dy=0 齐次微分方程！！！

麻烦大虾们帮个忙啊

Answer 1

令：v=y/x, y=xv, dy=vdx+xdv
dy/dx = -(y^2-2xy)/x^2
(vdx+xdv)/dx = 2v - v^2
v+xdv/dx = 2v - v^2
xdv/dx = v - v^2
dv/[v(1 - v)] = dx/x
∫dv/[v(1 - v)] = ∫dx/x
∫dv/v + ∫dv/(1 - v)] = ∫dx/x
lnv - ln(1-v) = lnx
ln(v/(1-v))=lnx+lnc
v/(1-v)=cx
(y/x)/(1-(y/x))=cx
y/(x-y)=cx
y=cx(x-y)

Answer 2

两边除以x^2((y/x)^2-2y/x)dx+dy=0令y/x=vdy=xdv+vdx(v^2-2v)dx+xdv+vdx=0(v^2-v)dx=-xdv1/xdx=1/(v-v^2)dv1/xdx=1/(v(1-v))dv=(1/(1-v)+1/v)dv两边积分得到：lnx=lnv-ln(1-v)+clnx=lny/x-ln((x-y)/x)+c此即为通解