xy'+y+xy^4=0
y=0是一解:
xy'/y^4+1/y³=-x;
齐次式:
xy'/y^4+1/y³=0
xy'+y=0
y'/y=-1/x
lny=-lnx+C1=ln(C2/x)
y=C2/x
变常数:
y'=C2'/x-C2/x²
代入:
xy'+y+xy^4=0
C2'-C2/x+C2/x+x(C2/x)^4=0
C2'+C2^4/x³=0
-C2'/C2^4=1/x³
(1/3)(1/C2³)=(-1/2)(1/x²)+C3
1/C2³=(-3/2)(1/x²)+3C3
C2³=1/[(-3/2)(1/x²)+3C3]
=2x²/(-3+6C3x²)
=2x²/(-3+C4x²)
C2=³√[2x²/(-3+C4x²)]
通解:
y=C2/x=³√[2x²/(-3+C4x²)]/x
=³√[2/(-3x+C4x³)]