工程數學problem~*

Question

Given that y = C1+ C2 X^2 is a two-parameter family of solutions

of xy"-y' = 0 on the interval(- 無限 , 無限) ,

Show that constants C1 and C2 cannot be found so that a member

of the family satisfies the initial conditions y(0)= 0 , y'(0)=1.

Explain why this does not violate the theorem.


Theorem :
Let An(x) , An-1(x) , ... ... ...A1(x) , A0(x) , and g(x) be continous on an
interval I , and let An(x)不等於 0 , for every X in this interval .
If X=Xo is any point in this interval , then a solution y(x) of the initial-
value problem y(Xo)=yo , y'(xo)=y1, ... ...[y^(n-1)]*(Xo)=y n-1 exists on
the interval and is unique.




我想知道這題的詳解答案~謝謝大大

Answer 1

(1)C1+C2*x²,都是xy''-y'=0之解

(2)
y(0)=0=C1
y'(0)=1=C2*0==>C2不存在
故ODE: xy''-y'=0沒有滿足初值條件y(0)=0,y'(0)=1之解

(3)
ODE:Ly=g(x)解存在且唯一的條件(簡單說)是
係數(函數)可微且在初值附近"不等於0"
本題初值給的是(0,0)處,而x=0時y''的係數x會是0,
故定理本身並未保證解存在且唯一

若本題的初值給y(1)=a, y'(1)=b (a, b任意),則必有解且唯一了

註:工程數學會探討理論性問題,不簡單喔!