高等数学:设有常微分方程？

Question

如图

Answer 1

（1）令x=u+h，y=v+k，则dx=du，dy=dv
(ax+by+c)/(dx+ey+f)=(au+ah+bv+bk+c)/(du+dh+ev+ek+f)
当ah+bk+c=dh+ek+f=0时
(ax+by+c)/(dx+ey+f)
=(au+bv)/(du+ev)
=(a+b*v/u)/(d+e*v/u)
φ[(ax+by+c)/(dx+ey+f)]
=φ[(a+b*v/u)/(d+e*v/u)]
=ξ(v/u)
dv/du=ξ(v/u)为齐次方程
此时，h=(bf-ce)/(ae-bd)，k=(cd-af)/(ae-bd)
（2）令p=ax+by，则dp/dx=a+b*dy/dx，
dy/dx=(dp/dx-a)/b=(1/b)*dp/dx-a/b
因为ae=bd，所以
(ax+by+c)/(dx+ey+f)
=b*(ax+by+c)/(bdx+bey+bf)
=b*(ax+by+c)/(aex+bey+bf)
=b*(p+c)/(ep+bf)
φ[(ax+by+c)/(dx+ey+f)]
=φ[b*(p+c)/(ep+bf)]
=η(p)
(1/b)*dp/dx-a/b=η(p)
dp/dx=b*η(p)+a=ε(p)
dp/ε(p)=dx为变量分离方程
（3）①dy/dx=(x+y+4)/(x-y-6)
令x=u+1，y=v-5
dv/du=(u+v)/(u-v)=(1+v/u)/(1-v/u)
令t=v/u，则v=tu，dv/du=u*dt/du+t

u*dt/du+t=(1+t)/(1-t)
u*dt/du=(1+t^2)/(1-t)
(1-t)/(1+t^2)dt=du/u
∫(1-t)/(1+t^2)dt=∫du/u
arctant-(1/2)*ln(1+t^2)=ln|u|+C
2arctant-ln(1+t^2)=ln(u^2)+C
[e^(2arctant)]/(1+t^2)=Cu^2
[e^(2arctan(v/u))]/(1+v^2/u^2)=Cu^2
e^[2arctan(v/u)]=C(u^2+v^2)
e^{2arctan[(y+5)/(x-1)]}=C(x^2-2x+y^2+10y+26)，其中C是任意常数
②dy/dx=(1+x-y)/(x-y)

令p=x-y，则dp/dx=1-dy/dx，dy/dx=1-dp/dx
1-dp/dx=(1+p)/p=1/p+1
dp/dx=-1/p
pdp=-dx
∫pdp=∫-dx
(1/2)*p^2=-x+C
p^2=-2x+C
(x-y)^2=-2x+C
x^2-2xy+y^2+2x=C，其中C是任意常数本回答被网友采纳