Diff equation for finite element method

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

Answer 1

[Mischa Kim]

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

Answer 2

[samuel]

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.