differential equation?
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
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
963003
int(y+2/2y^2+3y+1 dy)\n\n1/2*y^2 - 1/y + 3/2*y^2 + y + c\n\n2*y^2 - 1/y + y + c
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
750522
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
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
270795
so y= (x^(7/17))*((x-7)^2)\n  y'=((7/17)x^(7/17-1)) ((2)(x-7)^(2-1)(1))\n  y'=(7/(17x^(10/17)) (2(x-7))\n  Y'=(14x^(7/17)) (49/17^(10/17))\nand then\nset y'=0\nsolve for x.....
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
1130821
so for the first part, f(x)=x \nthe derivative no matter what is 1 \n\nfor the second part f(x)=1\nthe derivative no matter what is 0 \n\nso you will have combinational solution
803550
With respect to what?\n\nd/dx xy = y + x dy/dx\n\nd/dy xy = x + y dx/dy\n\nd/dt xy = x dy/dt + y dx/dt\n\nIt's just the Product Rule.
865054
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.
795130
y = tan(πx - 1)\nlet u = πx - 1\nnow, dy/dx = (dy/du)(du/dx)\nd(tan(u)/du = sec^2(u)\ndu/dx = π, so \ndy/dx = πsec^2(u)\ndy/dx = πsec^2(πx - 1)
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
1139627
dy/dx = 2x-3\nat X = 0. dy/dx = 2(0) -3 = -3
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
832532
dy/dx=cos x\n\nIntegrating both sides, we will get:\ny = sinx + c\n\ny(0) = sin0 + c = 0 => c=0\n\nSo y=sinx.