Finding solutions to equations

Question

Alexander Engman 2016-2-16

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/268361-finding-solutions-to-equations

评论： Torsten 2016-2-18

Hi!

I have a question regarding solving functions. I have a specific function I want to solve but rather than post it here, I will give a general description of my problem.

Lets say I have a function: eq1=integral from 0 to 2 of (k*x^2)/(0.2+k) and another function: eq2=(1+k)

The variable x represents the limits of the integral.

I want to solve eq1=eq2 for k>0, that is I want to find the exact value of the positive number k so that eq1=eq2.

My first thought was to put everything in a for-loop and let the loop generate values of k until it finds a value that satisfies the statement.

This is what I have so far:

x=0:2; %integral values for k=0:10000 %arbitrary positive k value eq1=(k*x^2)/(0.2+k) %first equation eq2=(1+k) %second equation if integral(eq1,0,2)==eq2 disp(k) end; end;

I am fairly certain that this method could work with a bit of effort but I don't know how to correctly express in the for loop that I want it to compare eq2 and the integral value of eq1 from x=2 to x=0.

Help appreciated!

Thanks.

PS. I am not sure that the two equations used in the example do have a solution for k, it was just used to show my problem.

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Answer 1

Torsten 2016-2-16

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https://ww2.mathworks.cn/matlabcentral/answers/268361-finding-solutions-to-equations#answer_209995

编辑：Torsten 2016-2-16

在 MATLAB Online 中打开

k0=1.0;
fun_int=@(x,k) k.*x.^2./(0.2+k);
fun=@(k) integral(@(x)fun_int(x,k),0,2,'ArrayValued',true)-(1+k);
k=fzero(fun,k0);

Best wishes

Torsten.

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Alexander Engman 2016-2-17

Could you please give me a brief explanation of how this code works?

Thanks!

Torsten 2016-2-18

在 MATLAB Online 中打开

%define starting guess for k
k0=1.0;                              
% define function to be integrated
fun_int=@(x,k) k.*x.^2./(0.2+k);   
% define function you want to equate to zero (eqn1-eqn2=0)
fun=@(k) integral(@(x)fun_int(x,k),0,2,'ArrayValued',true)-(1+k);    
% determine zero of function
k=fzero(fun,k0);