hello, im having problem trying to use the solve problem. it's not my first time using it but thata the first time its not working. i have this code line:
tmp = solve('x/L = (1-(h+eta)/ho) - A*log(((h+eta)/ho - A)/(1-A))','eta');
all the parameters are known except 'eta', still it does not provide a solution.
how can i fix it? any sugestions? thank you

Réponses (6)

Carlos
Carlos le 27 Mar 2013

1 vote

Using arbitrary values for x, h, h0, A and L
>> syms eta;
>> x=1;h=1;ho=1;A=0.1;L=1;
>> f=x/L -(1-(h+eta)/ho) - A*log(((h+eta)/ho - A)/(1-A));
>> eval(solve(f,eta))
ans =
-1.1068 - 0.2302i

1 commentaire

Jaimie Ritchie
Jaimie Ritchie le 22 Nov 2023
This was incredibly helpful to me in learning to use the solve function. I was able to plug in my equation and get a result. Thank you!

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Carlos
Carlos le 27 Mar 2013

0 votes

Doing
>> tmp = solve('x/L = (1-(h+eta)/ho) - A*log(((h+eta)/ho - A)/(1-A))','eta')% no semicolon here
tmp =
A*ho - h + (ho - A*ho)/(exp(wrightOmega(log((ho - A*ho)/(A*ho)) - (x/L + h/ho - (h - A*ho)/ho - 1)/A))*exp((x/L + h/ho - (h - A*ho)/ho - 1)/A))
It does give a solution.
Dany
Dany le 27 Mar 2013

0 votes

thats what im getting too. but i supose to have a numeric answer. i cant use the 'eval' function because it does not recognize 'wrightOmega'.
thats the problem.
Dany
Dany le 27 Mar 2013

0 votes

ok, thank you Carlos.
i'll give it a try

1 commentaire

Carlos
Carlos le 27 Mar 2013
Modifié(e) : Carlos le 27 Mar 2013
I think it should work, just remember to change the values of x,h... Please comment if something does not work by trying my approach to the problem.

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Dany
Dany le 27 Mar 2013

0 votes

sorry Carlos, it still does not work.
i try it with this values (real ones): h = 1.3721, ho = 500, L = 20000, x = 54.8828, A = -1.6452e-008.
and i still get the 'wrightOmega' and no answer .....
Dany
Dany le 27 Mar 2013

0 votes

ok it works..... thank you

1 commentaire

Carlos
Carlos le 27 Mar 2013
You are welcome, please mark my answer so people know the answer is correct and can use my answer for their problems.

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Question posée :

le 27 Mar 2013

Commenté :

le 22 Nov 2023

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