Recent content by highcoughdrop

http://www.physics.umd.edu/courses/Phys260/agashe/S09/solutions/HW11.pdf

http://www.physics.wisc.edu/undergrads/courses/spring08/202/hw2sols.pdf

After all of the mathematical shenanigans, I get A = B = C = 0 and D = 2. Therefore, y = 2sinx. and I never complained. lol

There are 4 equations to solve and they are : y(0) = 0 ---> A +B + C = 0 y(pi) = 0 ---> Ae^pi + Be^(-pi) - C = 0 y'(0) = 2 - -> A - B + D = 2 y''(0) = 0 ---> A + B-C = 0

The four solutions are 1, -1 -i and i.

Ahh, thank you. I think I knew this, I suppose I just need some clarification. I do believe that y = 2sinx is the only thing that is left after applying all of the intial conditions. Thank you much. =)

So far, my only success at this problem has been when y = 2 sin(x). And i believe that's the end of it. I was just looking for some mathematical way to solve the system.

Isnt there a systematic way of solving these? or is it just going to be a bunch of guesses?

It doesn't work because y(0) is never going to be zero, unless the coeffcient in front of it is 0, this y=0. However, usually this is a trivial solution, but in this case that doesn't even work because y'(0) is not = 2

Homework Statement (d4y)/(dx4) = y. Find the Unique solution y = y(x). Homework Equations Boundary Conditions: y(0)=0, y'(0) = 2, y''(0) = 0 , y(\pi) = 0 The Attempt at a Solution I really don't know where to start. I first started off with the guess that y = c1*sin(Ax) +...

Homework Statement An ice cube of mass 36 g at temperature -10 Celsius is placed in 360 g of water at 20 Celsius. Find thermal equilibrium temperature. \DeltaH(fusion) = 6.007 kJ mol\wedge-1. Molar heat capacity of ice = 38 water = 75 (Joules)/(K mol)...

Yeah, I'm aware of that, but the only reason I posted anything on here was so that the people who use google to search the physics forums have something when they do find this. Besides... that's how I got here, lol

Check this out, it may not help you much, but its something.

Alright. So, first of all, you should know that the moment of inertia of a rectangular plate, axis through the center of the plate is: I = (1/12)*M*(a^2+b^2) 1. Simply, for a square it would just be (a^2+a^2) or (2a^2) 2. Then, we find moment of inertia from the parallel-axis theorem...

Alright. So, first of all, you should know that the moment of inertia of a rectangular plate, axis through the center of the plate is: I = (1/12)*M*(a^2+b^2) 1. Simply, for a square it would just be (a^2+a^2) or (2a^2) 2. Then, we find moment of inertia from the parallel-axis theorem...