what is the solution to the equation x^y = y^x?
469518
There is an obvious 1-dimensional continuum of solutions to this equation: the line y=x.\nHowever, there are a continuum of more solutions, most of which are nontrivial:\nNote that log(x^y) = y*log(x). Thus log(x)/log(y) = x/y. This implies that y = x^(y/x) and similarly x = y^(x/y). This is the general form of the solution where the obvious solution is the restriction x/y = 1. \nIf we let x/y = 2, we get 2y = y^2, or y(y - 2) = 0 which gives nondegenerate solution point (4, 2) as mentioned previously. Setting x/y = 3 gives us 3y = y^3 or y^2 - 3 = 0, which gives the nontrivial solution point (3*Sqrt[3], Sqrt[3]) and so on.
824446
Not sure... but the answer is a complex number, I'll tell you that. There is no real solution for 5^x = x. In the equation a^x=x, a has to be less than e and greater than 0 for there to be a real solution.
931054
Group all elements with y together:\n\ny(x^2 + 6x +9) -x^2 + 2x + 3 = 0\n\nMove the rest to the other side:\n\ny(x+3)^2 = x^2 - 2x + 3 (1)\n\nx = -3 is obviously not the solution of this equation therefore we can divide both sides of (1) by (x+3)^2.\n\n(1) is equivalent to:\n\ny = (x^2 - 2x + 3) / (x+3)^2
518633
Call  Y of x like Y(x) \n\nY(x) = Y(x-1)+ (3*4^x)-1(1/9)x\nY(x) = Y(x-2)+ (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\nY(x) = Y(x-3)+ (3*4^(x-2))-1(1/9)(x-2) (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\n...\nY(x) = Y(1) + 3(4^x+...+4^2) - (1/9)(x+...+2)\nY(x) = Y(1) + 3*4[4^(x+1)-1]/[4-1] - [x(x+1)/2-1]/9\nY(x) = (sqrt3)/4 + 4^(x+2) - 4 - [x(x+1)/2-1]/9\nY(x) = 16 * 4^x - x(x+1)/18 - (37/9 - (sqrt3)/4)
700893
x^9-y^3z^12\n=(x^3)^3-(yz^4)^3 [x to the power 3 whole cubed minus yzto the power 4 whole cubed]\n=(x^3-yz^4) {(x^3)^2+(x^3)(yz^4)+(yz^4)^2}\n=(x^3-yz^4) (x^6+x^3yz^4+y^2z^8)
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
411840
x^2 - y^2 = (x+y)(x-y)\n\nso 25u^2-81= (5u+9)(5u-9)
744506
You have \nx^3 = x^2\nSubtract x^2 from each side\nx^3 - x^2 = 0\nFactor out x^2\nx^2 (x-1) = 0\nx^2 = 0 or x-1 = 0\nx = 0 or x = 1\n\nYou do not want to divide both sides by x^2 because then you will lose a solution.  Suppose I tried solving it this way:\nx^3 = x^2\nDivide both sides by x^2\nx^3/x^2 = x^2/x^2\nx = 1\nI only have one solution if I solve it this way.  The solution x=0 was lost when I divided by x^2\n\nNote that x^2 divided by x^3 is 1/x
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
465353
You must have meant y = z e^(-x). Substituting: \n\ny' = (z'-z)e^(-x)\ny'' = (z''-2z' + z)e^(-x)\n\ninto the equation gives (answer to (a)):\n\nz'' = -2\n\nwhich has solution (answer to (b)):\n\nz = -x^2 + Ax + B\n\nwhere A and B are constants, so the solution to the original equation is:\n\ny = (-x^2 + Ax + B)e^(-x)
707539
x = (y^2 + 2y -1) / 2\ndx=ydy+dy\n1=(y + 1 )dy/dx\ndy/dx=1/(y+1)\nd2y/dx2 = -1 / (y+1)^2
847176
f(x,y) = x^2*y*(e^(-x^2-y^2))\n\nfx(x,y) = 2x*y*e^(-x^2 - y^2)(1-2x^2) this is wrong:\nf_x(x,y) = x^2y ( e^(-x^2-y^2)) (-2x) + 2xye^(-x^2-y^2))\n= 2xye^(-x^2-y^2) [-x^2 + 1] =0\nif x=0 or y=0 or x^2 =1, \nx=0, y=0, x=1 or x=-1\n\nfy(x,y) = x^2*e^(-x^2 - y^2)(1-2y^2) \nf_y(x,y)=  x^2 y (e^(-x^2-y^2)) (-2y) +  x^2(e^(-x^2-y^2))\n=x^2e^(-x^2-y^2) [-2y^2 + 1] \nx=0, y=1/sqrt(2),y=-1/sqrt(2)\n\nso the critical points are: (0,y)\n(0,0), \n(1,1/sqrt(2)), (1,-1/sqrt(2))\n(-1,1/sqrt(2)),(-1,-1/sqrt(2))\ns
924424
The equations:\n(a+b)^2=a^2+2ab+b^2\nA^2-B^2=(A+B)(A-B)\nJames Chan's solutions were right.
187414
2^y = 64/y\n\n\nthis is the intersection of two curves:\nt=2^y\nand\nt=64/y\n\nthat is an exponential and an equilateral hyperbola.\nPlotting the graphs you can have an idea of the intersection (in this case you have the exact solution by plotting integer values, but in general you may need to solve by iterative numeric formulas, after finding graphically a rough estimate of the solution)
785428
y=e^cos(x) + cos(e^x)\ny'=-sin(x)e^cos(x) + -sin(e^x)e^x\n=-(sin(x)e^cos(x)+sin(e^x)e^x)
986257
If the equation is X^4 - 12x^2 -64 =0, there is.\n\nUse u = x^2, now your equation will look like this:\n\nu^2 - 12u - 64 =0\nFind u1 and u2 using the quadratic formula.\n\nThe solutions are u1 = 16, u2 = -4\n\nNow you go back to x, first deal with u1, then u2.  Because u = x^2 each u solution will give you two x solutions.\n\n\nu = x^2 = 16\nx = +/- SQRT(16)\nx1 = 4\nx2 = -4\n\nx2 = -4.  This is not possible with real numbers. so it will give you two imaginary numbers. Do the same as before\n\nx = +/- SQRT(-4) = +/- 2*SQRT(-1) = +/- 2i\n\nx3 = i\nx4 = -i
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.....