Diff equation for finite element method

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samuel
samuel on 20 Apr 2014
Answered: samuel on 21 Apr 2014
Hi all,
please could someone help me with this problem. I have to solve this diff equation y''= -6x+2 , x belongs to interval (0,2) and y(0)=y(2)=0
I tried to use dsolve
if true
x={0,2}
eqn2='D2y=-6(x)+2'
inits2='y(0)=0,y(2)=0'
y=dsolve(eqn2,inits2,'x')
end
but this does't work. And also tried to use ode45 function but i think this equation can't be solved by ode45 function. Could someone give me a hint how to do it, or better, example of code how to achieve solution. Many thanks
Sam

Answers (2)

Mischa Kim
Mischa Kim on 20 Apr 2014
Edited: Mischa Kim on 20 Apr 2014
Sam, use a bvp solver
function pdetest()
solinit = bvpinit(linspace(0,2,5),[0 0]);
sol = bvp4c(@mypde,@mybc,solinit);
x = linspace(0,2);
y = deval(sol,x);
plot(x,y(1,:));
end
function dydx = mypde(x,y)
dydx = [y(2); -6*x + 2];
end
function res = mybc(ya,yb)
res = [ya(1); yb(1)];
end

samuel
samuel on 21 Apr 2014
Hi Kim,
thanks a lot for your answer, sorry for long feedback, but this result is not suitable for me because i need result as an function, f.e. result of this equation should be
u(x)=x^2-x^3
and original qeuation is: -3y''=18x-6 U(x)=-2x this is how we can achieve this result (from my school book): because right side of diff eq is polynom of 1st degree so left side of diff eq is y(x) is polynom of 3rd degree -> y(x)=A + Bx + Cx^2 + Dx^3. So we'll solve coeficients A,B,C,D easily. If y(0)=0 then A=0, and when we appoint to diff eq left boundary point x=0 from interval (0,2) we get -3y''(0)= -6 -> -3*2C = -6, and from there C=1. When we appoint to diff eq right boundary point from interval (0,2) x=2 we get -3y''(2)=18*2-6 -> -3(2C+6D*2)=30, and from this we get D= -1. In the end we use condition 0=y(2)=A+B2+C*x^2+D^3, and from this we get B=2. And finally we get y(x)=2x+x^2+x^3. Soo final solution is u=U+y so
u(x)= -2x+[2x+x^2+x^3]=x^2+x^3
and this shoud be my output.
Pls Kim could you you help me how can I get this solution, because i am totally desperate
And this is solution for first element of finite element method.

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