1、x_{n+2}+6x_{n+1}+8x_{n}=2即(x_{n+2}+4x_{n+1}-2/3)+2(x_{n+1}+4x_{n}-2/3)=0令y_{n}=x_{n+1}+4x_{n}-2/3，则y_{n+1}=-2y_{n}解得：y_{n}=y_{1}*(-2)^{n}即x_{n+1}+4x_{n}-2/3=y_{1}*(-2)^{n-1}适当待定系数，可得x_{n+1}+(y_{1}/4)*(-2)^{n+1}-2/15=-4[x_{n}+(y_{1}/4)*(-2)^{n}-2/15]令z_{n}=x_{n}-(y_{1}/2)*(-2)^{n-1}-2/15则z_{n+1}=-4z_{n}解得：z_{n}=z_{1}*(-4)^{n-1}从而，x_{n}-(y_{1}/2)*(-2)^{n-1}-2/15=z_{1}*(-4)^{n-1}即：x_{n}=(y_{1}/2)*(-2)^{n-1}+z_{1}*(-4)^{n-1}+2/15而y_{1}=x_{2}+4x_{1}-2/3，z_{1}=x_{1}-(y_{1}/2)-2/15代入化简可得：x_{n}=(-4x_{1})[(-2)^{n-2}-(-4)^{n-2}]+(x_{2}/2)[(-2)^{n-1}-(-4)^{n-1}] -(1/3)*(-2)^{n-1}+(1/5)*(-4)^{n-1}-2/152、令y=u*e^{-x^{3}}是原方程的解代入化简可得u'-6u=1解得：u=Ae^{6x}-1/6从而原方程的解为：y=(Ae^{6x}-1/6)*e^{-x^{3}}
考数三，二阶微分，差分方程的通解特解形式需要熟记么？
