2. 考研
  3. 设y1(x)，y2(x)为y’+P(x)y=Q(x)的特解，又Py1(x)+2qy2(x)为y’+P(x)y=0的解，py1(x)-qy2(x)为y’+P(x)y=Q(x)的解，则p=_________，q=_________．

设y1(x)，y2(x)为y’+P(x)y=Q(x)的特解，又Py1(x)+2qy2(x)为y’+P(x)y=0的解，py1(x)-qy2(x)为y’+P(x)y=Q(x)的解，则p=_________，q=_________．

admin2019-09-04  29
问题 设y1(x)，y2(x)为y’+P(x)y=Q(x)的特解，又Py1(x)+2qy2(x)为y’+P(x)y=0的解，py1(x)-qy2(x)为y’+P(x)y=Q(x)的解，则p=_________，q=_________．
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答案[*]
解析 由一阶线性微分方程解的结构性质得

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考研数学三

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