Diophantiens eqn thrm:That if x0,y0 is on solution, then every solution is of the from:x=x0+(b/d)t;y=y0-(a/d)t
476679
First do this for d=gcd(a,b)=1. Subtracting the equations for the solutions (x_0,y_0) and (x,y), get that a(x-x_0)+b(y-y_0)=0. If gcd(a,b)=1, this shows that a divides y-y_0, i.e. y=y_0+at. Conclude that x=x_0+bt.\n\nNow see what happens if gcd(a,b)=1 and d is not 1. Then do the general case.
1174830
I don't think he asked for a solution of an eqn. but for factors\n\nx^4 + y^4 = (x^2 + y^2)^2  - 2.x^2.y^2\n\nRHS can be expressed as difference of 2 squares\nAnd so on.  I don't feel like doing the rest. you get the idea.
886265
gotta put the equation into y=mx+b first\ny= 1/3x - 1\n\nperpendicular would be y = -3x (+ or - any #)
900315
put,\ndy/dx = p\n\n\nHence,\ny = 2px + p^2\nor, p^2 + 2px - y = 0\nor, p = [-2x (+/-)(4x^2 + 4y)^(1/2)]/2\nor, p = [-x (+/-)(x^2 + y)^(1/2)]\nor, dy/dx = [-x (+/-)(x^2 + y)^(1/2)]\n\nIntegrate to solve.\nI am only showing the solution of, dy/dx= -x+(x^2+y)^(1/2)......(1)\n\nput, x^2 + y = Q^2\n\nHence, 2x + dy/dx = 2Q*dQ/dx,\nFrom (1) we have,\n2x + dy/dx= x+(x^2+y)^(1/2)\nor, 2Q * dQ/dx = x + p.............(2)\n\nNow put, Q = vx,\nhence dQ/dx = v + x * dv/dx,\nSubstituting in (2)\n2vx * (v + x * dv/dx) = x + vx\nor, 2v * (v + x * dv/dx) = 1 + v\nor, 2v^2 + 2vx dv/dx = 1 + v\nor,  2vx dv/dx = 1 + v - 2v^2\nor, 2vdv/(1 + v - 2v^2) = dx/x\nor, ∫2vdv/(1 + v - 2v^2) = ∫dx/x\nor, ∫2vdv/(1 - v)(1 + 2v) = ∫dx/x\nor, 2/3 *∫[(1 + 2v) - (1 -v)]dv/(1 - v)(1 + 2v) = ∫dx/x\nor, 2/3 *[∫dv/(1 - v) - ∫dv/(1 + 2v)] = ∫dx/x\nor, 2/3 [ - ln|1-v| - 1/2 ln|1 + 2v|] = ln|x| + c    \nor, 2/3 [ - ln|1-Q/x| - 1/2 ln|1 + 2Q/x|] = ln|x| + c \nor, 2/3 [ - ln|1-(x^2 + y)½/x| - 1/2 ln|1 + 2(x^2 + y)½/x|] = ln|x| + c  \n\n(x^2 + y)½ means (x^2 + y)^½\nPhewww............
814204
To quote the wikipedia:\n\n"Replace x by (x + y) / 2 and y by (x – y) / 2 in the product-to-sum formulae."
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
517123
It's hard to show it clearly here but the solution is actually:\n\n[-b±√(b²-4ac)]/2a\n\nthere will usually be two solutions!\nThe site below has it laid out more clearly...\n\n("invited" to leave school early.. 1974)
730660
This equation has a solution when x = 0 or y = 0\n\n- - - - - - - - - - - -\n\nx = 0\n\n5x - 28 = 12 + 3y\n\n5(0) - 28 = 12 + 3y\n\n0 - 28 = 12 + 3y\n\n- 28 = 12 + 3y\n\n- 28 - 12 + 12 + 3y - 12\n\n- 30 = 3y\n\n- 30/3 = 3y/3\n\n- 10 = y\n\nThe answer is y = - 10\n\n- - - - - - - - - - - - - - - -\n\ny = 0\n\n5x - 28 = 12 + 3y\n\n5x - 28 = 12 + 3(0)\n\n5x - 28 = 12 + 0\n\n5x - 28 + 28 = 12 + 28\n\n5x = 30\n\n5x/5 = 30/5\n\nx = 6\n\nThe answer is x = 6\n\n- - - - - - - - - \n\nx = 6\n\ny = - 10\n\n- - - - - - - - - -\n\nThe solution set is { 6, - 10 }\n\n- - - - - - - - - - - - -s-
744242
let y=4x(x+x^1/2)^1/2\nlny=ln4+lnx+1/2ln(x+x^1/2)\ny'/y=1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)\ny'=y[1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)]
982897
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
590565
y = mx + b\n\n0 = x + 3\n\ny can be anything, since it is not in the equation, and x will always be -3.
106013
y = -2x + 8\n\nsolve it yourself
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
133024
y = 0.\n\nETA: That's if we're to assume that the value of x is constant between the equations.
464039
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
787158
y = xtan(x)\n\nUse product rule:\ndy/dx = x/cos²(x) + tan(x)\n= x/cos²(x) + sin(x)/cos(x) \n= [x + sin(x)·cos(x)] / cos²(x)\n\nUse product and chain rules:\nd²y/dx² = 1/cos²(x) + (-2)·(x/[cos(x)]^3)·(-sin(x)) + 1/cos²(x)\n= 2/cos²(x) + [2·x·sin(x)] / [cos(x)]^3\n= [2·cos(x) + 2·x·sin(x)] / [cos(x)]^3
356024
x + y + z = t\n\nx +x% + y + y% + z + z% =\n[(100 + 1)( x + y + z)]% =\n101t% = t + t%. So\nx +x% + y + y% + z + z% = = t + t% or\nx% + y% + z% = t%