Solution Codes of Ordinary Diferenatial Equation
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I wrote MATLAB codes for 

equation with parameter n=450, a= -1.99 and x_0=3/4 as follows. 
clc; clear; clear all;
a=-1.99;  x0=3/4;
sumxx=[];
k=450;
for p=1:k
    T=linspace(p-1,p,101);
    sumx=1+a.*T;
    for n=0:p-2
        sumx=sumx+(a.^2).*(((1+a).^n).*(T-(n+1)));
    end
    sumx=sumx.*x0;
    sumxx=[sumxx,sumx];
    clear sumx
end
tt=linspace(0,k,k*101);
plot(tt,sumxx)
grid on
hold on

 But with the same parameters and in the equation above  at two places division by 4, i couldn't arrange MATLAB codes. Please, help me.
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回答(1 个)
  Jan
      
      
 2022-5-21
        
      编辑:Jan
      
      
 2022-5-21
  
      What is the purpose of the outer loop "for p=1:k"? Why does the inner loop "for n=0:p-2" run from 0 to p-2?
It is confusing that you've swapped n and k.
The purpose of your code is not clear to me. I have no idea, why inserting /4 twice is a problem and why you assume, that the graph is "not correct". I cannot guess what you expect instead.
n  = 450;
a  = -1.99;
x0 = 0.75;
T = linspace(0, 1000, 1000);
X = 1 + a * T;
for k = 0:n
    X = X + a^2 * (1 + a / 4)^k * (T - (k + 1) / 4);
end
X = X * x0;
figure;
plot(T, X)
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