Recent content by tkwan

It depends on how you have your resistors arranged. If you have two resistors in series with a source, then you will have the SAME current through both resistors. If your resistors are in series then you have different currents through them.

How did you get X0=1? If lambda=0, then X=A+Bx, using X'(0)=X(0), you get A=B and so X=A+Ax, using X'(L)=X(L), you get A=A+A(L), since L>0 A=0 and so X0=0? Am I missing something? Thanks.

I see what you did. Thanks a lot! I just ended up dividing both sides by B sin(uL) without thinking. Thanks. This leads me to a second question, for the lambda < 0, did I do that one right? I followed similar steps for that one. Thanks again. You're a real life saver.

I did try it and I got stuck. If I say X=Asin(ux)+Bsin(u(L-x)) we get X'=Aucos(ux)-Bucos(u(L-x)) and then I get stuck here. I tried both of my BC and couldn't get it to work. If I have X'(0)=X(0), I get Bsin(uL)=Au-Bucos(uL) and I don't know how to simplify that. If I use X'(L)=X(L), I get...

Sorry, I meant why should I

Thanks for the reply, but should I use the sin(ux), sin(u(L-x))? isn't the solution to this question cos and sin?

Homework Statement Consider the Heat Equation： du/dt=k(d2u/dx2), where d is a partial and d2 is the second partial. The B.C.'s are u_x(0,t)=u(0,t) and u_x(L,t)=u(L,t), where u_x is the partial of u with respect to x. The I.C is u(x,0)=f(x) Now, consider the Boundary Value Problem...