Use Laplace transforms to solve the following?
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
764897
use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
1139073
Differential Equations
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
432536
Oops………..!\n\ndac2chari\n\n\n u = 1/(ln t) Then du = t dt ………………..?  How that come …!\n\nIf u = 1/(ln t) then  u = [ ln t ] ^-1 \n\nAnd  du = - [ln t ] ^-2 [ 1/t ] dt = dt / [-t (ln t ) ^2] … please be careful.\n------------------------------------------------------------------------------------\n\nOK… this integral can’t carry out using analytical method; however, the flowing one can be solve\n\n∫ dt / [t ln(t)] \n\nu = ln (t)\ndu =  dt / t  \n\nGives\n\n∫ dt / [t ln(t)]  = ∫ du/ u  ln(u)\n\nSubstitute for u\n\n∫ dt / [t ln(t)]  = ln [ ln (t) ]
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
503650
Why don't you share with us what you have done so far instead of just asking someone to do your homework?\n\nSorry about the snide remark. Here is your well-deserved step-by-step solution:\nhttp://www.rgetso.com/yahoo/afchica101.pdf
1072090
There seems to be a typo in your function.  I'll assume you want the Laplace transform of:\n\n f(t)=t^2*exp(ln*5t)\n\n(If this is wrong, try reposting the correct function.)\n\nFirst, simplify the function\n\n f(t)= t^2*5t\n f(t) = 5*t^3\n\nThe Laplace transform of x^n, for integral n>0 is given by (see source):\n\nn!/(s^(n+1))\nso L[f(t)](s) = 5*3!/s^(4) = 30/s^4
787265
i find that the best method is the laplace transform..\nits quite simple..and you can find a table where each part of your differential equation will have a corresponding laplace (or 's') value..you can then simply solve and inverse it at the end to get back your result,\neg: dy/dt + dy^2/dt^2= 4u\nyou would convert to: s*Y(s) + s^2 * Y(s) = 4*U(s) [using the table]\n                                -> Y(s) [s+s^2] = 4*U(s)\n                                -> Y(s)             = ([4] / [s+s^2]) *U(s)\nnow depending on what u(t) is you find U(s). Eg. if u(t) is a step input, U(s) = 1/s.\n                                -> Y(s)             = ([4] / [s^2+s^3])\n                                -> y(t)               = 4.0*exp(-1.0*t)  [from the table]\n\nrefer to site: http://www.eecircle.com/applets/007/ILaplace.html\nfor an applet which does the calculations for you.\ngl
200344
W=f.delta(x) \nx=w/f = 49821/717= 69.49 metes
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
799788
You are given P(t=0), Q(t=0); P'(t=0), Q'(t=0); and P''(t=0), Q''(t=0). Given that P'' and Q'' remain the same (that is, the motion of P and Q do not change), you need to show that there is a time T such that P(t=T) = Q(t=T).\n\nIt's late and I'm drunk, so you can do the actual working for yourself :D
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
694227
Both Laplace and Fourier transforms are used for transformation from time-domain to freuency domain.\n\nThe basic difference between them can be understood from the transformation equations, We see that the Laplace transform can be viewed as a generalization of the Fourier transform from the real line (a simple frequency axis) to the entire complex plane. We say that the Fourier transform is obtained by evaluating the Laplace transform along the jw axis in the complex s plane. \n\nAn advantage of the Laplace transform is the ability to transform signals which have no Fourier transform.The Laplace transform can be seen as the Fourier transform of an exponentially windowed input signal
1015799
Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
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!
723707
s'(t) =(t-3)(-2t^{-3}) + t^{-2}\n=( -2t+6 +t)/ t^{-3}\n=(6-t) / t^{-3}\nso the velocity is 0 when the top =0\ni.e 6-t-0\nt=6\ns