2nd Order Nonlinear Differential Equation Solving with Central Difference Method?

I am trying to solve a 2nd order non linear differential equation using central finite difference method but ı cant, it is a boundary value problem
y''+2y'+5y=8sinx+4cosx
y(0)=0 and y(30)=0
step is dx=0.5
How can ı solve it, would you give me some suggestion about that or any script ?

답변 (3개)

Torsten
Torsten 2019년 3월 1일
http://web.mit.edu/10.001/Web/Course_Notes/Differential_Equations_Notes/node9.html
Furkan Celik
Furkan Celik 2019년 3월 3일
편집: Furkan Celik 2019년 3월 3일
Thank you so much for your response.. Im sharing script what i did.
clear all;
clc;
x = 0:3:30;
n = length(x);
y = zeros(1,n);
y(1,1)=0;
y(1,n)=0;
A = zeros(n-2);
B = zeros(1,n-2);
for i=1:n-2;
A(i,i) = 43;
end
for i=2:n-2;
A(i,i-1)=-2;
A(i-1,i)=4;
end
A
B(1,1)= 9*(8*sind(x(1,2))+4*cosd(x(1,2)));
B(1,n-2) = 9*(8*sind(x(1,n-1))+4*cosd(x(1,n-1)))
for i=2:n-3;
B(1,i)=9*(8*sind(x(1,i+1))+4*cosd(x(1,i+1)));
end
B
BB=B';
Y = inv(A)*BB;
YY=Y'
y(1,2:n-1)=YY(1,1:n-2)
plot(x,y);

댓글 수: 2

Can i ask a question?
how can you find A(i,i) A(i,i-1) and A(i-1,1)
Thank u
All explained on the above website.

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Thanawut Sumtib
Thanawut Sumtib 2021년 4월 5일
Use the finite difference method to solve y'' + 2 y' 3 y = 0 from t = 0 to 4 using a step size of 1 with y(0) = 1 and y’(0)=1.

카테고리

도움말 센터File Exchange에서 Numerical Integration and Differential Equations에 대해 자세히 알아보기

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2019년 3월 1일

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2021년 7월 17일

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