Question

微分方程(x)^2y'+2xy=3y^2,y(2)=2/9

Answer 1

方程两边同除以x^2,

dy/dx+2y/x=3(y/x)^2,

设u=y/x,

dy/dx=3u^2-2u

y=ux,

dy/dx=u+xdu/dx,

3u^2-2u=u+xdu/dx,

du/[3u(u-1)]=dx/x,

两边同时积分,

1/3∫[1/(u-1)-1/u]du=ln(Cx),(C是常数)

(1/3)[ln(u-1)/u]=ln(Cx)

(u-1)^(1/3)/u^(1/3)=Cx,

(y/x-1)^(1/3)/(y/x)^(1/3)=Cx,

通解为:[(y-x)/y]^(1/3)=Cx,

当x=2,y=2/9,

(2/9-2)^(1/3)/(2/9)^(1/3)=C*2,

-2=2C,

∴C=-1,

∴当y(2)=2/9特解是:

:[(y-x)/y]^(1/3)=-x.