What background information to I need to solve a seperable differential equation?
460400
You would need to know how to integrate (that is, find anti-derivatives). The proceedure forthis type of problem can usually be understood after first semester calculus. The specific integrations may need some second semester calculus.
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
74552
these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
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
1178962
To solve this it's a pretty straightforward application of Laplace transforms. You can do this question using convolutions as follows:\n\nFirst take the laplace transfom of the ODE:\n\nlet L{y(t)}(s) = Y(s) be the laplace transform of y(t).  Then we have:\ns^2(Y(s))  - 2s + 3Y(s) = 1/s^2 - (e^-s)/s\n\nrearranging we have Y(s) = (1/s^2 - e^-s/s + 2s) (1/(s^2+3)) = G(s) H(s) where G(s) = (1/s^2 - (e^-s)/s)  and H(s) =  (1/(s^2+3)) \n\nnow using the inverse laplace formula for the convolution of two functions:\n\ny(t) = L(^-1){Y(s)} = integral from 0 to t of (h(t-x)g(x)dx) + 2cos(sqrt(3)t)\n\n where h(t) = 1/sqrt(3) sin(sqrt(3)t) and g(t) = (t - u1(t))\n\nand the integral can be evaluated. \n\nSorry for the readability but I don't think yahoo answers can render math text properly. Also, as a note, this question would be far easier to solve using a partial fraction method rather than unecesarily using convolutions.
1109753
http://en.wikipedia.org/wiki/Lane-Emden_equation
437819
Actually, you get X''/X=C and Y''/Y=-C where C does not have to be 0. Also, remember that sums of solutions are solutions, so the solutions you get from separation of variables only give a basis for the vector space of solutions, not all the solutions. The specifics are difficult in this forum, but you have to set up the series and use the boundary conditions to find the coefficients of the series.
821342
THe VdW equation is P= nRT/(V-nb) - a(n/V)^2, just plug and chug.  But dont forget to switch your T into Kalvin
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
1051597
start with   d = m / v        multiply both side by v; \nyou get     v * d = m        divide both sides by d; \nyou get     v = m / d
1116002
partial differential equation (PDE) is a relation involving an unknown function of several independent variables and its partial derivatives with respect to those variables. Partial differential equations are used to formulate and solve problems that involve unknown functions of several variables, such as the propagation of sound or heat, electrostatics, electrodynamics, fluid flow, elasticity, or more generally any process that is distributed in space, or distributed in space and time. Very different physical problems may have identical mathematical formulations.
526667
First of all, the characteristic equation is m^3-3m^2+3m-1=0, which factors as (m-1)^3=0, for m=1 a triply repreated root.\n\nThis gives e^x, xe^x, and x^2 e^x as three independent solutions for the *homogeneous* equation (the one with a 0 on the right hand side). Now you have to find a *particular* solution of the non-homogeneous equation. The problem is that e^x is already a solution of the homogeneous version. So now you go up in the heierarchy to try a particular solution of the form\ny_p=Ax^3 e^x.\nPlug this into the original equation to solve for A. Then add this particular solution to the general solution of the homogeneous part to get your general solution. Check to be sure there are three arbitrary constants!
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!
806346
using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know
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) ]
976124
dy/dx = (dy/dt)/(dx/dt)\ndy/dt = 1/(2√t)\ndx/dt = 1\n\ndy/dx = 1/(2√t)\nx=t\ny=√t\n\nSubstituting t=1/4:\n\ndy/dx = 1\nx = 1/4\ny = 1/2\n\nUsing point-slope form:\n\ny-1/2 = 1(x-1/4)\n\nSimplifying:\n\ny=x+1/4
438946
A way to estimate a derivative and define a general way of solving differential equations numerically. See the links for more information.