how to use the 'solve' function?

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Dany
Dany on 27 Mar 2013
Commented: Jaimie Ritchie on 22 Nov 2023
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

Answers (6)

Carlos
Carlos on 27 Mar 2013
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 Comment
Jaimie Ritchie
Jaimie Ritchie on 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 on 27 Mar 2013
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 on 27 Mar 2013
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 on 27 Mar 2013
ok, thank you Carlos.
i'll give it a try
  1 Comment
Carlos
Carlos on 27 Mar 2013
Edited: Carlos on 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 on 27 Mar 2013
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 on 27 Mar 2013
ok it works..... thank you
  1 Comment
Carlos
Carlos on 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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