what is "one dimensional Burgre equation" with intial & boundary conditions?
457083
I think you mean Burger's equation.  Go to any good book on partial differential equations and you will see it there.  Richard Haberman has a good book for PDE's.
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
969305
This initial value problem is surprisingly simple, since niether y nor any derivatives other than the second appear in the equation, so all that is necessary is to integrate twice:\n∫∫d²y/dx² dx dx = ∫∫6x dx dx\n∫dy/dx dx = ∫3x² + C dx\ny=x³+Cx+K\n\nNow we need merely find the constants C and K. We know that y'(0)=3(0)² + C = 0, so C=0. Further, we know that y(0) = K = 1, so K=1.\n\nThus y=x³+1\n\nb: There is exactly one such function. All antiderivatives of a function are the same up to addition of an arbitrary constant, and as we showed when solving the problem, there is only one consistent set of values for the two constants of integration introduced in this problem.
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
101802
I don't think there is a unique solution for it. You have more unknown variables than the equations.
1107625
I assume you are able to take partial derivatives. So take the partial derivatives that appear in the heat equation (you did look up the heat equation, right?) and plug them in to see if you get equality.\n\nSo, the partial derivative with respect to t is\n-lambda^2 k*u(x,t). This is the same as k times the second partial derivative with respect to x. Looks right to me!
712970
dy/dx=2xy\ndy/y=2xdx\nintegrate\nlny=x^2+C\nput in initial conditions\nln(1)=1+C\nC=0-1\n'e' each side\ny=e^(x^2-1)\n\ncheck in original equation\ndy/dx=2x*e^(x^2-1)\n2x*y=2x*e^(x^2-1)\nthis answer is correct\nat x=1.6\ny=e^(1.6^2-1)=4.758821245\n\nto apply euler's method\nto the initial value problem\ndy/dx=f(x,y),  y(x0)=y0\nproceed as follows\n(a)take x0 and Y0=y0 as starting\nvalues,choose a step size h,\nand set r=0\n(b)set x(r+1)=r+h\n(c)calculate an approximation \nY(r+1) to(y(r+1)),using the\niteration formula \nY(r+1)=Yr+hf(xr,Yr)\n(d)if furtherapproximate values\nare required,increase r by 1,and\nreturn to sub-step (b)\n\ni also have details of the \nclassical forth order runge-\nkutta method,but will keep \nthat for a later date\n\ni hope that this helps
98111
Code:\nx'' - 2nd derivative of x with respect to time (t) (acceleration in x)\n\nx' - 1st derivative of x wrt time (velocity in x)\n\nAnd the same system for the y coordinate.\n\nx''=1.2 then by integration wrt time gives x'=1.2t+constant. Boundary conditions are x'=4750 m/s at t=0, so the constant = 4750 m/s. Equation is now Vx = x' = 1.2t + 4750. Substitute in t=743s, answer is Vx=5641 m/s. \n\nSame method for y'' with acceleration of 8.5 m/s^2, and boundary conditions of y'=0 at t=0 gives Vy=6315.5 m/s.\n\nFinal answers:\nVx=5641.6 m/s\nVy=6315.5 m/s
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) ]
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
982897
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
15744
Classes of Solutions to Mathieu's Equation\n\n  Floquet's theorem asserts that any solution to the Mathieu equation (Eq. gif) is of the form:[4,29,28]\n\n \n\nwhere is either real or imaginary and is a -periodic functiongif of . This assertion implies that solutions of Eq. gif can be separated into the product of a term of the form and a series of harmonics, , each of which is -periodic in . For values of with a non-zero imaginary component and , Eq. gif indicates that:\n\nThus, , yields an unbounded solution to the Mathieu equation. On the other hand, values of which are purely real yield bounded, oscillatory solutions for .\n\nThere are two types of bounded solutions to Mathieu's equation: functions of fractional order and functions of integral order. Functions of fractional order are obtained for . The oscillatory periods of the frequencies associated with fractional-order Mathieu functions are non-integer multiples of . The special case of corresponds to an -periodic (in terms of ) solution. These -periodic solutions are Mathieu functions of integral order.\n\nAll bounded solutions to the Mathieu equation--- those of fractional as well as integral order--- are described (in accordance with Floquet theory) by an infinite series of harmonic oscillations whose amplitudes decrease with increasing frequency:\n\n \n\nThe argument of in Eq. gif represents harmonics of frequency , while the term represents a (secular) frequency which is a function of the trap specifications (or, equivalently, of the Mathieu parameters a, q). The Mathieu parameters and also determine the coefficients , which approach zero as n increases.[36]
110904
No. you may count yourself lucky to have these conditions because with these you are able to get an unique solution ψ.\nBut there are 4 main conditions:\n\nNorm: |ψ| := 1\n\nContinuity of ψ\n\nContinuity of ψ'    (only for finite potentials)\n\nLimit: ψ(x) -> 0 for x -> ∞
247965
infinite values for both,\nits caus u cant solve two variables with just one equation
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
785253
So are you looking for a function with a singularity on (0,1)?  That would address the question I think.\nmaybe (x-(1/2))^(1/2)\n\nTHE FIRST ANSWER IS WRONG, I THINK, BECAUSE YOUR DOMAIN IS OPEN AND DOESN'T INCLUDE 0.
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember