i need help in "Laplacian differential operator" (Read details)...Level :2nd year university Physics...
492128
I'm a third year physics student and I know what a Laplacian differential operator is, I just don't understand your question. Could you clarify?
276224
these guys do it much better than i can:\n\nhttp://hyperphysics.phy-astr.gsu.edu/Hbase/class/phscilab/kundt2.html
353539
For recent physics and maths preprints, try here\n\nhttp://lanl.arxiv.org/\n\nor one of it's mirror sites
1139073
Differential Equations
450143
log into www.wikipedia.org and search for lie group...there is a detailed description and all the formulas as well and its free.
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
894731
Here are a few Laplace Transforms you need to know:\n\nL[f(x)] = F(s)\nL[f'(x)] = s*F(s) - f(0)\nL[2e^2x] = 2/(s-2)\n\nNow substitute these values into your differential equation:\n\ns*F(s) - f(0) + 9(F(s)) = 2/(s-2).  Plug in f(0) = 0 and solve for F(s):\n\nF(s)[s+9] = 2/(s-2) ------------> F(s) = 2/[(s-2)(s+9)]\n\nNow use the inverse Laplace transform to solve for f(x).  To do this, you need to transform F(s) into partial fractions.  Partial fractions state that:\n\n2/[(s-2)(s+9)] = A/(s-2) + B/(s+9) where A and B are constants.  Multiply both sides by the denominator (s-2)(s+9)\n\n2 = A(s+9) + B(s-2) ------->2 = As + 9A + Bs - 2B\n\nNow combine like terms to create equations to solve for A&B\n\n2 = s(A+B) + (9A-2B)\n\ns terms -------------------> A+B=0 (on the left hand side, there are no 's' terms, that's why the equation equals to zero)\nconstant terms ---------> 9A-2B = 2\n\nUsing these two equations, A = 2/11, B = -2/11\n\nThus,\n\n2/[(s-2)(s+9)] = 2/[11(s-2)] - 2/[11(s+9)]\n\nNow use the inverse Laplace transform:\n\nSince the L^-1[1/(s-a)] = e^(at), we can solve for f(x):\n\nf(x) = (2/11)(e^[2t]) - (2/11)(e^[-9t])\n\n------------\n\nBelow is a website to help you with Laplace transforms.  Hope this helps
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
643352
The result is still under debate because all transformations have still not been verified. The Sobolev space can be bi variant if the gauge is not satisfied.\n\nI would suggest looking up the bulletins of the London Mathematical Society, their late 2004 editions came out with one such treatment.
83265
That's an elliptic integral.  Check out the link.
925536
To solve, first, you need to know certain Laplace Transforms.  You can find them from a table or solve for them using the definition:\n\nL[f'(x)] ---------> sF(s) - f(0)\nL[f''(x)] --------> s^2F(s) - sf(0) - f'(0)\nL[1] ------------> 1/s\nL[x] ------------> 1/s^2\n\nPlugging this into the equation, you get:\n\n[s^2F(s) - sf(0) - f'(0)] - [sF(s) - f(0)] = 1/s - 1/s^2\n\nPlugging in the initial conditions and simplifying the right side, you get:\n\ns(s-1)*F(s) = (s-1)/s^2 ---------------> Now solve for F(s)\n\nF(s) = 1/s^3\n\nNow use the general inverse Laplace Transform property:\n\nL-1[1/s^n] ---------> x^(n-1)/(n-1)!\n\nIn our case, n = 3, pluggin in you get:\n\nf(x) = x^2/2\n\n--------------\n\nHope this helps
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
756366
Look this one up on the internet.  I doubt you're gonna find anyone here to do it for you.  That's too much information.  Don't you have a textbook? Use that.
786148
Electronic engineering - circuit analysis:\n\nSolving a circuit for the current and voltage according to Kirchhoff's Laws and these device equations:\nResistor   : v = R * i\nCapacitor: i = C * dv/dt\nInductor   : v = L * di/dt\n\nAnything complicated you generally solve by using the Laplace transform, in which case the above equations simplify to\nfor V,I = Laplace transforms of v,i:\nResistor   : V = RI\nCapacitor: I = sCV\nInductor   : V = sL\n\nThen you solve the system of equations for V,I and apply a reverse Laplace transform.
421380
from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck!
149751
As far as I am aware, the solutions manual is only available to instructors.  I am, however, aware that David Pace (administrator of DavidPace.com) used to have solutions for a large number of problems on his website.  Unfortunatly, he has recently taken these down without significant comment on the reason for doing so, so they are somewhat unavailable.  You can, however, access a cached version of them from google if you really wish to attempt to locate them.\n\nI encourage you, however, to struggle your way through each and every problem in electrodynamics before ever checking to see if you arrived at the correct result.  It really is the only way to learn.\n\nAlternatively, if you're having significant difficulty, you might try posting to Yahoo! answers in the physics section and I'm sure that someone would be interested in helping you get over your difficulty.\n\nI wish you good luck in your study of electrodynamics using one of the better books out there.  Good luck with your exam.
758976
The total differential is [∂F/∂x]dx +[ ∂F/∂y]dy\n\n∂F/∂x = 4x^3*y^-1\n∂F/∂y = x^4*(-y^-2) = -x^4y^-2