solve laplace eq'n:(d^2V/dx^2)+(d^2V/dy^2)=0.?
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.
1139073
Differential Equations
279194
It easily follows that the general solution to this differential equation is y=Ae^{kt} for some constant A [You may differentiate to double check your answers.] \n\nSo given the initial condition, i.e. y(t=60)=0.5*A, we obtain k=(ln(0.5))/(60).
1081914
(dx/dy) = (xy^3) / (2y^4 + x^4)\n\nLet x = y*u.\nThen (dx/dy) = u + y*(du/dy).\n\nSubstituting these in the given equation we have,\n\nu + y*(du/dy) = (u*y^4) / {2y^4 + (u*y)^4} = u/(2 + u^4)\n\nor y*(du/dy) = u/(2 + u^4) - u = -u*(1 + u^4)/(2 + u^4)\n\nor ∫ [-(2 + u^4)/{u*(1 + u^4)}] du = ∫ [1/y] dy _________(1)\n\n\n\nNow, let us first express [-(2 + u^4)/{u*(1 + u^4)}] in easily integrable form.\n\n-(2 + u^4)/{u*(1 + u^4)} \n\n= -{2*(1 + u^4) - u^4}/{u*(1 + u^4)}\n\n= -{2*(1 + u^4)/{u*(1 + u^4)} + u^4}/{u*(1 + u^4)}\n\n= -2/u + u^3/(1 + u^4)\n\nTherefore, ∫ [-(2 + u^4)/{u*(1 + u^4)}] du\n\n= ∫ [ -2/u + u^3/(1 + u^4)] du\n\n= -2∫ [1/u] du + ∫ [ u^3/(1 + u^4)] du \n\n= -2*ln(u) + (1/4)*ln(1 + u^4) + k\n\nwhere k = arbitrary constant of integration\n\nHence, from (1) we have \n\n-2*ln(u) + (1/4)*ln(1 + u^4) + k = ln(y)\n\nor  (1/4)*ln(1 + u^4) + k = ln(y) + 2*ln(u) = ln(y) + ln(u^2) = ln(y*u^2)\n\nor k = ln(y*u^2) - (1/4)*ln(1 + u^4) \n\n= ln(y*u^2) - ln[(1 + u^4)^(1/4)]\n\n= ln[(y*u^2)/{(1 + u^4)^(1/4)}]\n\n\nNow, substituting u = x/y we have\n\nk = ln[(x^2/y)/{(1 + (x/y)^4)^(1/4)}]\n\nor k = ln[(x^2)/{(x^4 + y^4)^(1/4)}]\n\nLet k = ln(c).\n\nThen c = (x^2)/{(x^4 + y^4)^(1/4)}\n\nor x^2 = c*(x^4 + y^4)^(1/4)
697597
1. Solving for dy/dx gives dy/dx = -y/(x+y), but this doesn't fit the form of the Bernoulli equation.  Instead, flip it around and get dx/dy + (1/y)x = -1, so n=0, P(y)=1/y, and Q(y)=-1.  In this case, there's no need for a change of variable, so instead just jump directly to using the integrating factor m(y) = e^(integral of 1/y) = y, to get\n\ny dx/dy + x = -y, so (xy)' = -y.  Therefore\n\nx = (1/y) (-y^2/2 + C), or x = -y/2 + C/y.\n\nMultiply through by 2y: 2xy + y^2 = 2C, which is equivalent to the solution you expect, since C is an arbitrary constant.\n\n2. Since n=2, let v = y^(1-n) = 1/y.  After transforming, you get \n\ndv/dx + vp(x) = q(x)\n\nwhere p(x) = (1-n)P(x) = 1/x and q(x) = (1-n)Q(x) = x, so you're solving\n\ndv/dx + (1/x)v = x.\n\nNow use the integrating factor x, and  you have\n\nx dv/dx + v = x^2\n\nso (vx)' = x^2, or v = (1/x)(x^3/3 + C), so\n\nv = x^2 / 3 + C/x.  But v = 1/y, so \n\ny = 1 / (x^2/3 + C/x) = x / (x^3/3 + C).\n\nThis doesn't agree with what you're saying the answer should be, but it does satisfy the differential equation (as I confirmed by plugging in this solution), whereas your answer doesn't.
255082
by laplace\n   s^2f(s)-sf(0)+4f(s)-4f(0)+9f(s)=9
894731
Here are a few Laplace Transforms you need to know:\n\nL[f(x)] = F(s)\nL[f'(x)] = s*F(s) - f(0)\nL[2e^2x] = 2/(s-2)\n\nNow substitute these values into your differential equation:\n\ns*F(s) - f(0) + 9(F(s)) = 2/(s-2).  Plug in f(0) = 0 and solve for F(s):\n\nF(s)[s+9] = 2/(s-2) ------------> F(s) = 2/[(s-2)(s+9)]\n\nNow use the inverse Laplace transform to solve for f(x).  To do this, you need to transform F(s) into partial fractions.  Partial fractions state that:\n\n2/[(s-2)(s+9)] = A/(s-2) + B/(s+9) where A and B are constants.  Multiply both sides by the denominator (s-2)(s+9)\n\n2 = A(s+9) + B(s-2) ------->2 = As + 9A + Bs - 2B\n\nNow combine like terms to create equations to solve for A&B\n\n2 = s(A+B) + (9A-2B)\n\ns terms -------------------> A+B=0 (on the left hand side, there are no 's' terms, that's why the equation equals to zero)\nconstant terms ---------> 9A-2B = 2\n\nUsing these two equations, A = 2/11, B = -2/11\n\nThus,\n\n2/[(s-2)(s+9)] = 2/[11(s-2)] - 2/[11(s+9)]\n\nNow use the inverse Laplace transform:\n\nSince the L^-1[1/(s-a)] = e^(at), we can solve for f(x):\n\nf(x) = (2/11)(e^[2t]) - (2/11)(e^[-9t])\n\n------------\n\nBelow is a website to help you with Laplace transforms.  Hope this helps
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
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
821024
∫cos(x^2) dx = √(π/2)FresnelC(√(2/π)x)
492313
x^2*y"-x*y' +4*x*e^(x^3)=0\n\nx[x*y"-y' +4*e^(x^3)]=0\n\n=> x=0 or x*y"-y' +4*e^(x^3)=0\n\nx*y"-y' +4*e^(x^3)=0\n\n=>\n\n∫[x*y"-y' +4*e^(x^3)].dx=0\n\n=> [-4*x*Gamma[1/3, -(x^3)]]/[3(-(x^3))^(1/3)]=0\n\nwell even my computer couldn't solve that, so let the answer be x=0.
220918
2*x^2 - 3*x - 2 = 0 <=> x = -1/2 V x = 2 \n\n==> 2*x^2 - 3*x - 2 <=0 <=> x = [-1/2 , 2]
349210
dy/dx = -(15cos(3/x)/2x^2) * (sin(3/x))^3/2
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
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.
940656
dy/dx = -(e^x + e^(-x)) / -(y (e^x - e^(-x) + 1)³)\n\nSo ∫ydy = ∫(e^x + e^(-x)) dx/(e^x - e^(-x) + 1)³ \n\nLet u = e^x - e^(-x) + 1 thus du = (e^x + e^(-x) ) dx\n\nSo ∫ydy = ∫du/u³\n\nSo ½y² = -½*1/u² + ½A\n\nie y² = A - 1/u²\n= A - 1/(e^x - e^(-x) + 1)²
802168
The function is of the form d(u/v) which is given by\n(v du - u dv) / v^2.\nhere u=(1+x)^2, hence du=2(1+x)\nhere v=1-x, hence dv=-1.\nsubstitute these values in the formula and simplify to get final result.\n\nKSR