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

[AiRAVATA]

Well, in fact is easy to come up with such solution.

The general solution for $y''(x)+y(x)=0$ is $y(x)=A\cos x+ B\sin x$. Evaluating the boundary conditions, the constants $A,\,B$ are determined. After doing that, one can conclude that $A=0$ and $B=\cos^{-1}(1)$.

 

[jt316]

am not sure how to apply this type of boundary condition

 

[jt316]

If y(x) = Acos(x) + Bsin(x)

y'(x) = Bcos(x) - Asin(x)

y''(x) = -Acos(x) - Bsin(x)

y(1)= - y(-1)
y'(1) = 2-y'(-1)
but I'm still not sure about this type of boundary condition...

 

[AiRAVATA]

Just evaluate:

$$y(-1)+y(1)=A\cos(-1)+B\sin(-1)+A\cos(1)+B\sin(-1)=2A\cos(1)=0,$$

so $A=0$. What is $B$?

 

[jt316]

I see what I'm doing wrong. I was trying to put the boundary conditions into y(x) and y'(x) instead of puting those two into the boundary conditions.

B= 1/cos(1)

Thanks for the help

oh yeah [I hope it's the last question :) ] but how did you find the general solution to y''(x) + y(x)=0 to be y(x)= Acos(x) + Bsin(x) ?

 

[AiRAVATA]

Well, you propose (like in any linear second order ode with constant coefficients) an exponential solution $y(x)=e^{rx}$, and then find out what is the value of $r$, i.e.

$$y''(x)+y(x)=(r^2+1)e^{rx}=0.$$

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

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.

 

[jt316]

I'll try to find that book. Thanks again for the help.