Solve the general solution of a Differential equation with boundary conditions

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Vlasis Demosthenes
Vlasis Demosthenes 2018 年 11 月 29 日
コメント済み: Torsten 2018 年 11 月 29 日
Hi
I have the general solution of a differential equation and the boundary conditions. How can I solve this?
w(x)=C1+C2*x+C3*(x^2)+C4*(x^3)+(L^4)*C5*cosh(x/L)+(L^4)*C6*sinh(x/L)
w(0)=0
w'(0)=0
w''(0)=0
w''(L)=0
w'''(L)-(L^2)*w'''''(L)=F/(E*I)
w''(L)-(L^2)*w''''(L)

採用された回答

Torsten
Torsten 2018 年 11 月 29 日
Tell the program which gave you the general solution to additionally incorporate the six boundary conditions.
  2 件のコメント
Vlasis Demosthenes
Vlasis Demosthenes 2018 年 11 月 29 日
I don't have the exact differential equation from which the general solution emerged.
Is there any way to solve this? from the general solution and the boundary conditions only?
Torsten
Torsten 2018 年 11 月 29 日
Build w', w'', w''' and w'''' of w and evaluate the conditions. You'll arrive at a linear system of equations in the Ci's as unknowns.
I'll do it for the first two conditions:
w(0)=0 leads to
C1+C2*0+C3*(0^2)+C4*(0^3)+(L^4)*C5*cosh(0/L)+(L^4)*C6*sinh(0/L) = 0, thus
C1 + L^4*C5 = 0
w'(0)=0 leads to
C2+2*C3*0+3*C4*(0^2)+(L^4)/L*C5*sinh(0/L)+(L^4)/L*C6*cosh(0/L) = 0, thus
C2 + L^3*C6 = 0
...
Once you have the equations, write them as a system of linear equations in the form A*C = b with a 6x6 matrix A and a 6x1 vector b.
Then C is given by
C=A\b
Best wishes
Torsten.

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