Question

求dy/dx=1/(2x-y^2)的解

Answer 1

dx/dy=2x-y^2,
dx/dy-2x=-y^2,(1)
令dx/dy-2x=0,
dx/x=2dy,
lnx=2y+lnC1,
x=C1e^(2y),
用参数变易法,用v代替C1,
x=ve^(2y),(2)
dx/dy=e^(2y)dv/dy+2ve^(2y),(3)
把（2）（3）式代入（1）式,
e^(2y)dv/dy+2ve^(2y)-2ve^(2y)=-y^2,
dv=-y^2dy/e^(2y),
v=-∫y^2dy/e^(2y),
=（-1/2）[y^2e^(-2y)+ye^(-2y)+(1/2)e^(-2y)}+C2
代入（2）式,
x=(-1/2)e^(2y){[y^2e^(-2y)+ye^(-2y)+(1/2)e^(-2y)}+C2}
=(-1/2)[y^2+y+1/2+Ce^(2y)].