一阶微分方程y'=2((y+2)/(x+y-1))^2
1个回答

设m=y+2,n=x-3

∴代入原方程得dm/dn=2[m/(m+n)]²

==>dm/dn=2/(1+n/m)².(1)

再设n/m=t,则dn=tdm+mdt

∴代入方程(1)得dm/(tdm+mdt)=2/(1+t)²

==>t+mdt/dm=(1+t)²/2

==>mdt/dm=(1+t²)/2

==>2dt/(1+t²)=dm/m

==>2arctant=ln│m│-ln│C│ (C是积分常数)

==>e^(2arctant)=m/C

==>m=Ce^(2arctan(n/m))

==>y+2=Ce^(2arctan((x-3)/(y+2))

故原方程的通解是y+2=Ce^(2arctan((x-3)/(y+2)) （C是积分常数）.

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