dy/dx-2y/(x+1)=(x+1)^2

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dy/dx-2y/(x+1)=(x+1)^2
急求、、请好心人发下过程.阿门
求解微分方程。。

∵dy/dx-2y/(x+1)=0 ==>dy/y=2dx/(x+1)
==>ln|y|=2ln|x+1|+ln|C| (C是积分常数)
==>y=C/(x+1)²
∴设原方程的通解为 y=C(x)/(x+1)² (C(x)表示关于x的函数)
代入原方程整理得 C'(x)=(x+1)^4 ==>C(x)=1/5(x+1)^5+C (C是积分常数)
==>y=(x+1)³/5+C/(x+1)²
故原方程的通解是 y=(x+1)³/5+C/(x+1)² (C是积分常数)

dy/dx,y=(1+x+x^2)e^x dy/dx-2y/(x+1)=(x+1)^2 dy/dx-2y/(1+x)=(x+1)^3 dy/dx=(e^x+x)(1+y^2)通解 dy/dx=x(1+y^2)/y通解 y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy (x^2)dy+（y^2)dx=dx-dy dy/dx-y/x=x^2 从(dx)/(dy)=1/y '导出：(d^2x)/(dy^2)=-y''/(y')^3 y=1(2x+1) 求dy/dx 求解微分方程dy/dx+x/2y=1/2 dy/dx=1+x+y^2+xy^2