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)一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)一qy2(x)为y′+P(x)y=Q(x)的解，则p=__________，q=____________．

admin2019-01-05  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)一qy2(x)为y′+P(x)y=Q(x)的解，则p=__________，q=____________．
选项
答案由一阶线性微分方程解的结构性质得 [*]
解析
转载请注明原文地址:https://kaotiyun.com/show/V0W4777K
0

考研数学三

相关试题推荐
随机试题
最新回复(0)