In general, it is the orthogonality condition of eigenvalue problems. That is why I expect an identity matrix. Thanks,

Dear Prof. Mariner: Yes, that is it. I have a question about "assume" command. When should I use the assume command? Thank you very much.

Dear Prof: The result is almost what I want to get. However, I do not know why it has "I" which is imaginery symbol, I think. restart; invlaplace(-1/(s*(s^2+a^2)),s,t); As a trial, MAPLE gave me "cosine" function in the result. All of parameters are constant and positive. Is it possible to obtaine the result without "I" symbol? Thank you

Thanks anyway, I am trying to solve it for time variables. Could you make some code for arbitrary constants of omega and r for time t[2] and t[3] solutions? Those should be in positive format due to time variables. I think that you have to use "solve" command. Thanks

Thank you for your numeric way. But, I try to get a solution symbolically. t[1],t[2],...t[5] are time variables. oemga, r are constants. Could you give me help? Thanks,

Dear Dr. Alec: Which one is the book that you wrote? Thanks

Dear Dr. Alec: Which one is the book that you wrote? Thanks

Thanks a lot, Dr. Alec. However, I try to get the same answer as yours by setting up the equations. Here is my code. I think that s2 should be the same as s1. Please tell me why it does not give me the same results. Thank you Sung restart: t[1] := 0:t[2] := delta:t[4] := t[3]+delta:t[6] := t[5]+delta: u1 := t[2]-t[1]+t[4]-t[3]+t[6]-t[5] = 1/2*U: delta := solve(u1, delta): eq1 := expand(sum((-1)^(i+1)*cos(omega*t[i]), i = 1 .. 6) = 0): eq2 := expand(sum((-1)^(i+1)*sin(omega*t[i]), i = 1 .. 6) = 0): eq3 := expand(sum((-1)^(i+1)*cos(r*omega*t[i]), i = 1 .. 6) = 0): eq4 := expand(sum((-1)^(i+1)*sin(r*omega*t[i]), i = 1 .. 6) = 0): _EnvAllSolutions := true: _EnvExplicit := true: s1:=solve({eq1, eq2, eq3, eq4}, {r, t[3], t[5]}); trs1 := simplify(eq1*sin(omega*delta)+eq2*(1-cos(omega*delta))): trs2 := simplify(eq3*sin(r*omega*delta)+eq4*(1-cos(r*omega*delta))): rs1 := factor(trs1): rs2 := factor(trs2): s2:=solve({rs2, rs1}, {r, t[3], t[5]});

In SOLVE command, Why do you include the variable r? Is there any way to use the following command instead of your command? s1: = solve({eq1,eq2,eq3,eq4},{t[3],t[5]}); Could you explain why you use "r" variable which is positive real? Thank you in Smoking Section. Y-G Sung