求一道微分方程,y'=（x+y）^3这一类的题不是很懂,麻烦讲的详细些
1个回答
设u=x+y,则dy/dx=du/dx-1
代入原方程
得du/dx-1=u³ ==>du/dx=u³+1
==>du/(u³+1)=dx
==>1/3[1/(u+1)+(2-u)/(u²-u+1)]du=dx
==>2/3[2/(u+1)-(2u-1)/(u²-u+1)+3/(u²-u+1)]du=dx
==>2/3[2/(u+1)-(2u-1)/(u²-u+1)+4/(1+((2u-1)/√3)²)]du=dx
==>2/3[2ln|u+1|-ln|u²-u+1|+2√3arctan((2u-1)/√3)]=x+C/3 (C是积分常数)
==>2/3[ln|(u+1)²/(u²-u+1)|+2√3arctan((2u-1)/√3)]=x+C/3
==>ln[(x+y+1)^4/(x²+2xy+y²-x-y+1)²]+4√3arctan[(2x+2y-1)/√3]=3x+C
故原方程的通解是
ln[(x+y+1)^4/(x²+2xy+y²-x-y+1)²]+4√3arctan[(2x+2y-1)/√3]=3x+C (C是积分常数)
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