Is anyone familiar with LaPlace Transforms and how to solve them?
969006
I've done a good bit with linear transforms in general (LaPlace,Fourier, Z), so hopefully I can help you. The most important thing for your class is to know whether or not they will allow the use of tables or not. Even the simplest function will usually require integration by parts to compute a transform directly. Usually I use tables of transforms coupled with properties of transforms (like the differentiation property, which lets you use algebra instead of calculus to solve an ordinary linear DE). I really can't go into any depth without pencil and paper. Check out the link below (it contains an example of using LaPlace transforms to solve a linear ordinary DE).
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
267169
dP/P(a-bP)=dt\n\nusing partial fraction to integrate the left side of the eq.\n\nSince 1/P(a-bP)=(1/a)/P + (b/a)/(a-bP)..then\n\ndP/[(1/aP)+(b/a)(1/(a-bP))]=dt ...... Integrate \n\n(1/a)Ln(P)-(1/a)Ln(a-bP)=t+c   ...... c is a constant\n\n(1/a) Ln[P/(a-bP)]= t+c\nOR\nLn[P/(a-bP)]= at +c1 ....... c1=a*c\nOR\nP/(a-bP) =exp[at+c1]\nP/(a-bP) =exp(at)*exp(c1) ...... let exp(c1)=c2\nP/(a-bP) = c2 exp(at)\n\nBut P=f(t) then\n\n1/[c2 exp(at)] = (a-bP)/P\n-->\n(1/c2) exp(-at)=(a/P)-b  .... let 1/c2 = c3\n-->\nc3 exp(-at)+b= a/P\n-->\nP(t)= a/[c3 exp(-at)+b]\nOR\n            a\nP(t)= ---------------     ..... where c3 is a constant\n      c3 exp(-at) + b
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Detailed information as well as equations are available on the following page : \n\nhttp://www.jwave.vt.edu/crcd/kriz/modules/module06/formulation/edge_formulation.html
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Part 1:\n\nWe can rewrite y^x as (e^ln(y))^x which is e^(xln(y))\n\n∂f/∂x f(x,y)= ∂f/∂x (x^y) + ∂f/∂x(y^x)\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)e^(xln(y))\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)y^x\n\n\nPart 2: There is no simple function which is the integral of e^(x^2).
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Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
1178575
In mathematics, the Laplace transform is a powerful technique for analyzing linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems, to name just a few. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.\n\nThe Laplace transform is an important concept from the branch of mathematics called functional analysis.\n\nIn actual physical systems the Laplace transform is often interpreted as a transformation from the time-domain point of view, in which inputs and outputs are understood as functions of time, to the frequency-domain point of view, where the same inputs and outputs are seen as functions of complex angular frequency, or radians per unit time. This transformation not only provides a fundamentally different way to understand the behavior of the system, but it also drastically reduces the complexity of the mathematical calculations required to analyze the system.\n\nThe Laplace transform has many important applications in physics, optics, electrical engineering, control engineering, signal processing, and probability theory.\n\nThe Laplace transform is named in honor of mathematician and astronomer Pierre-Simon Laplace, who used the transform in his work on probability theory. The transform was discovered originally by Leonhard Euler, the prolific eighteenth-century Swiss mathematician.\n\nDefinition:\nThe Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s).\n\n\n\nThe parameter s is in general complex:\n\n \nThis integral transform has a number of properties that make it useful for analysing linear dynamical systems. The most significant advantage is that differentiation and integration become multiplication and division, respectively, with s. (This is similar to the way that logarithms change an operation of multiplication of numbers to addition of their logarithms.) This changes integral equations and differential equations to polynomial equations, which are much easier to solve.\nhttp://en.wikipedia.org/wiki/Laplace_transform\n\nThe Laplace transform is a powerful tool formulated to solve a wide variety of initial-value problems. The strategy is to transform the difficult differential equations into simple algebra problems where solutions can be easily obtained. One then applies the Inverse Laplace transform to retrieve the solutions of the original problems\nhttp://www.efunda.com/math/laplace_transform/index.cfm\n\nAlso try these ...\nhttp://www.sosmath.com/diffeq/laplace/basic/basic.html
414104
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
796927
try the link below...... hope it can put an end to ur integral problems
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
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dP=[ aP^2 - bP ] dt\n\nintegrate from the initial point t0 to ant time t\n\nP=[ a/3 P^3 - b/2 P^2] {0 t}\nsubs B= aP^2  D=bP\nP=[1/3 * B * P - 1/2 D * P] {0 t}\n\nP - P0 = 1/3 * (B-B0)* (P-P0) - 1/2 (D-D0) * (P-P0)\n\ndivide by (P-P0)\n\n1 = 1/3 (B-B0) - 1/2 (D-D0)\n6 = 2B - 2B0 -3D + 3D0\n6 + 2B0 -3D0 = 2aP^2- 3bP\n\nM= 2*(3bP)*P/3*(2aP^2)=6b/^a= b/a\n\nThe RHS : D0P0/B0= (bP0*P0)/(aP0^2)= b/a\n\ntherefore M= RHS
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I think the numerical recipes book is the gold standard for solutions to these and many other kinds for problems.\n\nThis book is freely available online at\n\nhttp://www.library.cornell.edu/nr/cbookcpdf.html\n\nFor your questions take a look at chapter 10\n\nIf you can master this book you should have no shortage of programming opportunities.
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CONFORMAL MAPPING\nIn a limited group of problems one can use a short cut to the solution of the\nLaplace's equation, conformal mapping. The problem has to be 2D (two\ndimensional), i.e., the boundaries, boundary conditions and sources contain\nonly two variables, x and y say.\nSome examples where this method can be used is: the electrostatic potential;\nheat conduction; flow of fluids.\n\nThe conformal mapping relies on the properties of analytic functions. \n\nAny function that has continuous partial derivatives of the second order and\nthat satisfies the Laplace's equation is called a Harmonic function.\nThus both the real part, u, and imaginary part, v, of f are harmonic\nfunctions.\nIf the function f = u+iv is analytic, u and v are conjugate harmonic\nfunctions. Given one of two harmonic functions, the Cauchy-Riemann\nequations can be used to find the other.\nThat u and v are harmonic functions can be used to find the potential. Both\nu and v are candidates for the potential since they satisfy Laplace's\nequation. Alternatively the function f can be viewed as a change of\nvariables, a transformation from the complex z-plane to the complex w-\nplane.\nTransformation of boundary conditions.\nTwo types of boundary conditions remain unchanged by the transformation\nand problems with these boundary conditions are best suited for conformal\nmapping. The types are:\n1) The value of a harmonic function is constant on the boundary.\n(one type of Dirichlet problem)\n2) The normal derivative of a harmonic function is zero on the boundary.\n(one type of Neumann problem)\n(The conditions are that f and its inverse are analytical)\nThe capacitance between two conductors\nIf Q is the electrical charge on either conductor and D V is the\ndifference in potential between one conductor and the other, then the\ncapacitance C is defined to be c=q/v
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put,\ndy/dx = p\n\n\nHence,\ny = 2px + p^2\nor, p^2 + 2px - y = 0\nor, p = [-2x (+/-)(4x^2 + 4y)^(1/2)]/2\nor, p = [-x (+/-)(x^2 + y)^(1/2)]\nor, dy/dx = [-x (+/-)(x^2 + y)^(1/2)]\n\nIntegrate to solve.\nI am only showing the solution of, dy/dx= -x+(x^2+y)^(1/2)......(1)\n\nput, x^2 + y = Q^2\n\nHence, 2x + dy/dx = 2Q*dQ/dx,\nFrom (1) we have,\n2x + dy/dx= x+(x^2+y)^(1/2)\nor, 2Q * dQ/dx = x + p.............(2)\n\nNow put, Q = vx,\nhence dQ/dx = v + x * dv/dx,\nSubstituting in (2)\n2vx * (v + x * dv/dx) = x + vx\nor, 2v * (v + x * dv/dx) = 1 + v\nor, 2v^2 + 2vx dv/dx = 1 + v\nor,  2vx dv/dx = 1 + v - 2v^2\nor, 2vdv/(1 + v - 2v^2) = dx/x\nor, ∫2vdv/(1 + v - 2v^2) = ∫dx/x\nor, ∫2vdv/(1 - v)(1 + 2v) = ∫dx/x\nor, 2/3 *∫[(1 + 2v) - (1 -v)]dv/(1 - v)(1 + 2v) = ∫dx/x\nor, 2/3 *[∫dv/(1 - v) - ∫dv/(1 + 2v)] = ∫dx/x\nor, 2/3 [ - ln|1-v| - 1/2 ln|1 + 2v|] = ln|x| + c    \nor, 2/3 [ - ln|1-Q/x| - 1/2 ln|1 + 2Q/x|] = ln|x| + c \nor, 2/3 [ - ln|1-(x^2 + y)½/x| - 1/2 ln|1 + 2(x^2 + y)½/x|] = ln|x| + c  \n\n(x^2 + y)½ means (x^2 + y)^½\nPhewww............
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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.)
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This is a separable differential equation\ndP/dt = 4P^2 + tP^2\nFactor P^2 out from the left side\ndP/dt = p^2(4+t)\nDivide each side by P^2 and multiply each side by dt\ndp/P^2 = (4+t)dt\nNow integrate each side\n-1/P = 4t + (1/2)t^2 + C\nSolve for P and you will get\nP = -1/(4t + (1/2)t^2 + C)