solving 4 equation with four unknwon

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mohamed sayed
mohamed sayed 2016 年 2 月 16 日
編集済み: John BG 2016 年 2 月 16 日
i can't find the sol for this code . what's rong with it ?
syms k m n d
i=.015; o=.016; x=[0.1;0.2;0.3;0.4;0.5]; v=-104.63; t=97;
equ1=d-.5*k*m*n;
equ2=n-m-(2*t)/k;
equ3=(k/4)*(n^2-m^2+i^2-o^2)+d*log((m*o)/(n*i))-t*(m-i-o+n);
equ4=(.25)*(i^2)*x-(-1/v)*((t/3)*(i^3-m^3+o^3-n^3)+.125*k*(i^4-m^4+n^4-o^4)+.5*d*(m^2-i^2+o^2-n^2)); sol=solve(equ1,equ2,equ3,equ4)

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John BG
John BG 2016 年 2 月 16 日
編集済み: John BG 2016 年 2 月 16 日
introducing your system in muPAD, as you keyed it in the answer, returns
Warning: 8 equations in 4 variables.
simplifying it to:
  • d-.5*k*m*n=0
  • n-m-10/k=0
  • k*(n^2-m^2)+d*log(m/n)-(m+n)=0
  • 1+(m^3+n^3)+k*(n^4-m^4)+d*(m^2-n^2)=0
muPAD return null, kind of can't find answer.
However, if you approximate log(x) (natural or Neper log) by only 1st Taylor series ln(x)~x-1, bear in mind that evaluating ln(x0)~x-x0-1 , then
  • d-.5*k*m*n=0
  • n-m-10/k=0
  • k*(n^2-m^2)+d*(m/n)-1-(m+n)=0
  • 1+(m^3+n^3)+k*(n^4-m^4)+d*(m^2-n^2)=0
muPAD returns
Use attached muPAD result file if you want to carry on from here.
If you find this answer of any help solving your question please click on thumbs-up vote above link, thanks in advance
John
note: d*log(m/n) should be d*m/n-d, not d*m/n-1

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