在常微分方程y''+p(x)y'+q(x)y=0中在方程y''+p(x)y'+q(x)y=0中,如果p(x),q(x)在(-无穷,+无穷)上连续,那么它的任一非零解在xoy平面上不能与x轴相切.为什么?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 23:49:51
![在常微分方程y''+p(x)y'+q(x)y=0中在方程y''+p(x)y'+q(x)y=0中,如果p(x),q(x)在(-无穷,+无穷)上连续,那么它的任一非零解在xoy平面上不能与x轴相切.为什么?](/uploads/image/z/13379068-28-8.jpg?t=%E5%9C%A8%E5%B8%B8%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8By%27%27%2Bp%28x%29y%27%2Bq%28x%29y%3D0%E4%B8%AD%E5%9C%A8%E6%96%B9%E7%A8%8By%27%27%2Bp%28x%29y%27%2Bq%28x%29y%3D0%E4%B8%AD%2C%E5%A6%82%E6%9E%9Cp%28x%29%2Cq%28x%29%E5%9C%A8%28-%E6%97%A0%E7%A9%B7%2C%2B%E6%97%A0%E7%A9%B7%29%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E9%82%A3%E4%B9%88%E5%AE%83%E7%9A%84%E4%BB%BB%E4%B8%80%E9%9D%9E%E9%9B%B6%E8%A7%A3%E5%9C%A8xoy%E5%B9%B3%E9%9D%A2%E4%B8%8A%E4%B8%8D%E8%83%BD%E4%B8%8Ex%E8%BD%B4%E7%9B%B8%E5%88%87.%E4%B8%BA%E4%BB%80%E4%B9%88%3F)
在常微分方程y''+p(x)y'+q(x)y=0中在方程y''+p(x)y'+q(x)y=0中,如果p(x),q(x)在(-无穷,+无穷)上连续,那么它的任一非零解在xoy平面上不能与x轴相切.为什么?
在常微分方程y''+p(x)y'+q(x)y=0中
在方程y''+p(x)y'+q(x)y=0中,如果p(x),q(x)在(-无穷,+无穷)上连续,那么它的任一非零解在xoy平面上不能与x轴相切.为什么?
在常微分方程y''+p(x)y'+q(x)y=0中在方程y''+p(x)y'+q(x)y=0中,如果p(x),q(x)在(-无穷,+无穷)上连续,那么它的任一非零解在xoy平面上不能与x轴相切.为什么?
在方程y''+p(x)y'+q(x)y=0中,如果p(x),q(x)在(-无穷,+无穷)上连续,那么它的任一非零解在xoy平面上不能与x轴相切.
求y'=0,A(n,0)y₁=x代入y''+P(x)y'+Q(x)y=0得P(x)+xQ(x)=0.(1)
y₂=xlnx代入y''+P(x)y'+Q(x)y=0得1/x+P(x)(lnx+1)+Q(x)(xlnx)=0.(2)
(1)(2)联立得P(x)=-1/x,Q(x)=1/x²
通解为:y=C1x+C2xlnx(C1,C2为任意实数)【不能】