f(x)=Pn(x)e型 设方程y"+py+q=Pn(x)e特解形式为*=O(x)e,则得 Q(x)+(24+)Q(x)+(2+p+q)Q(x)=Pn(x)-(大) 提示 y*+py*+qy*-[Q(x)e]+[Qx)eay+qQ(x)e Q"(x)+2Q(x)+2Q(x)]ex+pQ(x)+x)ex+qQ(x)ex [Q"(x)+(2A+p)g(x)+(+p+q)Q(x)]lx 首页上页返回下 结束 铃首页 上页 返回 下页 结束 铃 一、 f(x)=Pm (x)e x 型 提示 =[Q(x)+(2+p)Q(x)+( 2+p+q)Q(x)]e x  =[Q(x)+2Q(x)+ 2Q(x)]e x+p[Q(x)+Q(x)]e x+qQ(x)e x y*=Q(x)e x 设方程y+py+qy=P  m (x)e x 特解形式为 下页 Q(x)+(2+p)Q(x)+( 2+p+q)Q(x)=Pm (x) ——(＊) 则得 =[Q(x)e x ]+[Q(x)e x ]+q[Q(x)e x y*+py*+qy* ]