求y'-2/(X+1)y=(x+2)^3的通解

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求y'-2/(X+1)y=(x+2)^3的通解
求y'-2/(X+1)y=(x+2)^3的通解

求y'-2/(X+1)y=(x+2)^3的通解
求y'-[2/(x+1)]y=(x+2)³的通解
先求齐次方程y'-[2/[(x+1)]y=0的通解.
分离变量得dy/y=2dx/(x+1)；积分之得lny=2ln(x+1)+lnC₁=lnC₁(x+1)²；
故得y=C₁(x+1)²； 将C₁换成x的函数u,得y=u(x+1)².(1)
将(1)的两边对x取导数得dy/dx=2u(x+1)+(x+1)²(du/dx).(2)
将(1)和(2)代入原方程得2u(x+1)+(x+1)²(du/dx)-[2/(x+1)][u(x+1)²]=(x+2)³
化简得(x+1)²(du/dx)=(x+2)³
故得du/dx=(x+2)³/(x+1)²；分离变量得du=[(x+2)³/(x+1)²]dx=[(x+4)+(3x+4)/(x+1)²]dx；
积分之得u=∫(x+4)dx+∫[(3x+4)/(x+1)²]dx=(x+4)²/2+3∫xdx/(x+1)²+4∫d(x+1)/(x+1)²
=(x+4)²/2+(3/2)∫d(x+1)²/(x+1)²-3∫dx/(x+1)²+4∫d(x+1)/(x+1)²
=(x+4)²/2+(3/2)ln(x+1)²-3∫d(x+1)/(x+1)²+4∫d(x+1)/(x+1)²
=(x+4)²/2+(3/2)ln(x+1)²+∫d(x+1)/(x+1)²=(x+4)²/2+(3/2)ln(x+1)²-1/(x+1)+C
代入(1)式即得原方程的通解为y=(x+1)²[(x+4)²/2+3ln∣x+1∣-1/(x+1)+C]

从方程的形式来看P=-2/(X+1)     Q=(x+2)^3

根据公式

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