Solving Diff.Eq. with Boundary Conditions: y(x) = x

In summary, the given equation y(x)=x+(1/2)*∫(from u=-1 to 1)[(1-|x-u|)y(u)du], x є [-1, 1] is a solution of y''(x)+y(x)=0 with the given boundary conditions of y(1)+y(-1)=0 and y'(1)+y'(-1)=2. By using the Fundamental Theorem of Calculus, the derivative of y(x) is found to be y'(x)=1+(1/2)*[f(x,x)+∫(from u=-1 to x)[∂f/∂x(u,x)du-g(x,x)+
  • #36
Well, you propose (like in any linear second order ode with constant coefficients) an exponential solution [itex]y(x)=e^{rx}[/itex], and then find out what is the value of [itex]r[/itex], i.e.

[tex]y''(x)+y(x)=(r^2+1)e^{rx}=0.[/tex]

That way, [itex]r=\pm i[/itex] and the solution is [itex]y(x)=Ae^{ix}+Be^{-ix}[/itex]. As you only want real solutions, then [itex]y(x)=C\cos x+D\sin x[/itex].

You should really check a book on ODE's. I recommend you the one written by Boyce and DiPrima Elementary Differential Equations. It should be in your library.
 
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  • #37
I'll try to find that book. Thanks again for the help.
 
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