solve for laplace equation .. square..?
439125
Doesn't a constant V=Vo work? You don't even have to separate variables for this one! Remember the solution is unique, so if you find one, you have found them all.\n\nThe coefficients for the non-constant terms of your series will be 0, but you forgot the constant term. That corresponds to k=0.
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
1009002
y = x² - 4x - 5\n\nSimplify -5 = 4 - 9 to complete the square\ny = x² - 4x + 4 - 9\n\nSimplify the square\ny = (x - 2)² - 9\n\n^_^
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
1117258
Area of a square = L * L\n\nArea of A = x * x ..........(1)\nArea of B = y *y  ..........(2)\n  \n y = x -1               ..........(3)\n\nA = 4 * B              ...........(4)\n\nSubstitute (1) & (2) into (4) gives:\n\nx^2 = 4 y^2           ...........(5)\n\nSub. (3) into (5)\n\nx^2 = 4 (x-1) ^2\n\nSolve:\n\nx^2 = 4( x^2 -2x +1)\nx^2 = 4x^2 -8x +4\nOr\n\n3x^2 - 8x +4 =0\n\nSecond order eq.\n\nx1,x2= [+8 +- Sqr(8^2 - 4*3*4)]/(2*3)\n\nx1,x2 = [8+-4]/6\n\nx1 = 2   &  x2 = 4/6\n\nThen\ny1 = 2-1 =1   & y2 = 4/6 -1 = -2/6 ( illogical )\n\nthen\n\nx= 2  and y = 1
821024
∫cos(x^2) dx = √(π/2)FresnelC(√(2/π)x)
766270
im assumuing you mean a rectangle? If it was a square itd be 400, if its a rectangle its 80l-l^2
445638
l=2w, and length times width is area, so...\nl*w=A...use l=2w, w=w, A=5000\n2w*w=5000, or\n2w^2=5000...divide by two and solve for w!
539943
x+y........e=mc2..........2+2
694690
determine the square of 6 - sqrt[(x-2)^2 + y^2]\n\n-->\n\n36 + [(x-2)^2 + y^2] - 12*sqrt[(x-2)^2 + y^2]\n= 40 + x^2 + 2x + y^2 - 12*sqrt[(x-2)^2 + y^2]
243669
divde the area by pi, then take the square root.  Multiply that reult by 2 and then by pi and your done.
880796
dy/dx = √(y/x) = √y/√x\nFirst of all, group √x,dx and √y,dy to get:\n(1/√y)dy = (1/√x)dx \ny^(-1/2)dy = x^(-1/2)dx\n\n∫ y^(-1/2)dy = ∫ x^(-1/2)dx\n2y^(1/2) = 2x^(1/2) + C1\ny^(1/2) = x^(1/2) + C\ny = (√x + C)²
934968
Okay, assuming you mean:\n∫e^(2x)/√(1-e^(2x)) dx\nLet u=1-e^(2x), du=-2e^(2x) dx, dx = -1/2 e^(-2x)\n-1/2 ∫1/√u du\n-1/2 ∫u^(-1/2) du\n-u^(1/2) + C\n-√(1-e^(2x)) +C
815728
first this is not a PDE but ODE\n\nsimple integration would do
946504
Intensity varies as the square of the distance.\nThat should help
779215
(-4x^4)² = (-4x^4) * (-4 x^4)\n= (- 4) *( -4) * (x^4) * (x^4)\n= +16 * x^(4 + 4)\n= 16x^8
140062
L x W = Area\n\nCheater