mathematical model using differential equation?
111123
Change in temperature with time is = constant * the difference in temperature. T' = k*(T2-T1)\nso: T' /T2-T1 = k\nintegrating\nln(T) = kt +C\nT = exp(kt+c) = C* exp(kt) {exp is e,2.71  to the exponent}\nI heated up a cup of water and measure the time for the temperature to drop about half.\nSolved for k\nThen measured how long the microwave took to heat and plugged that in and found out the microwave is equililant to 600 F. I think I made a mistake, this seems too low.\nSee if you can do better.
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Mathematical model are widely used in agriculture in or der to determine variation of populations of plants or animals.\n\nYou have a link  in source
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
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∫_0_2 g(t) dt = 5\n\n(a)\ny= ∫_0_4 g(t/2) dt\nlet t/2 =x ; so dt = 2dx; and the range becomes x:0 to 2\nso, y= ∫_0_2 g(x) 2 dx = 2  ∫_0_2 g(x) dx = 2 * 5 = 10\n\n(b)\ny= ∫_0_2 g(2-t) dt\nlet 2-t=x; so dt = -dx; range becomes x:2 to 0\ny= - - ∫_0_2 g(x) dx = 5
887510
Differential equations.\nLinear algebra.\nNumerical analysis.
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Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
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You want to start by integrating the equation. Fortunately, this is not difficult.\n\ndy/dx = (x-2)e^(-2y)\n\nInt(e^(2y)dy) = Int((x-2)dx)\n\n (1/2)e^(2y) = (1/2)x^2 - 2x + C, where C is some constant\n\ne^(2y) = x^2 - 4x + C  Note that e^(-2y) = 1/[x^2 - 4x + C]\n\n2y = ln (x^2 - 4x + C)\n\ny = (1/2)ln (x^2 - 4x + C)\n\nWe are almost done, now we just need to plug in the given condition, and solve for C.\n\nln (2) = (1/2)ln(4 - 8 + C)\n2ln(2) = ln(C-4)\n4 = C - 4\nC = 8\n\nSo the particular solution is y = (1/2)ln (x^2 - 4x + 8)\n\nIf you want to check this, we need to differentiate it\n\ndy/dx = (1/2)(2x-4)/(x^2 - 4x + 8)\n\nRemember from above that  e^(-2y) = 1/[x^2 - 4x + 8]\n\nSo dy/dx = (x-2)e^(-2y) as required
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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
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D =M/V\n\nM=DV\n\nV=M/D
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Use Charles' Law\n\nLet V initial = V\nLet T initial = T\n\nLet V final = V'\nLet T final = T'\n\n\nV/T = V'/T'\n\nSolve for V' you might have to convert from mL to L though.
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rate of change=dy/dx=slope
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use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
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A first-order differential equation of the form:\n\na1(x)*(dy/dx) + a0(x)*y = g(x)\n\nis a linear equation.\n\nWhen g(x) = 0, it's homogenous.\n\nSo, (x^2-9)(dy/dx) + xy = 3 is linear and similarly, (x^2-9)(dy/dx) + xy = 0 is homogenous.\n\nA first-order differntial equation of the form\n\ndy/dx = g(x)h(y)\n\nis separable.\n\nSo dy/dx = (y^2)x(e^(3x+4y)) is seperable since (y^2)x(e^(3x+4y)) = (x(e^3x))((y^2)(e^4y)), and dy/dx = y + sin(x) is NOT separble since it cannot be defined as a product of two equations seperately of x and y.\n\nI can't recall how to discern if a diff equation is exact, but this webpage\n\nhttp://www.utpb.edu/scimath/wkfield/mod2/Exact.html\n\ngives a very detailed explanation.\n\nHope that helps.
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use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
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Begin with where you left of:\n\n2x + 2yy' = 0\n\nDifferentiate again, just like the first time.  Use the chain rule on the yy' term:\n\n2 + 2(y'y' + yy'') = 0\n1 + y'^2 + yy'' = 0\ny'' = (-1-y'^2)/ y\n\nFrom your y' = -x/y\n\ny'' = (-1-(-x/y)^2)/ y = -x^2/y^3 - 1/y\n\nThis isn't the answer you have but see if they are the same:\n\nx^2 = 4 - y^2\n\ny'' = -(4-y^2)/y^3 - 1/y = -4/y^3 + 1/y - 1/y = -4/y^3\n\nSo the two solutions are the same
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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).
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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) ]