solve the exact differential equation (-1 x^4+1 y)dx+(1x-4x^2)dy=0 Solution Let

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Let x and y be the two numbers, where y>x. Two numbers differ by 3: If y>x and x and y differ by 3, y-x =3 difference between their reciprocals is 4: If y>x, then 1/x > 1/y, so (1/x) - (1/y) = 4 Solve the 2nd eqn for x: (1/x) - (1/y) = 4 1/x = 4 + (1/y) x = 1/ (4 + (1/y)) Substitute for x into 1st eqn: y - 1/ (4 + (1/y)) = 3 -1/ (4 + (1/y)) = 3 - y 4 + (1/y) = 1/(3-y) 4 = 1/(3-y) - (1/y) Use y(3-y) as common denominator 4 = y/(y(3-y)) - (3-y)/(y(3-y)) 4 = (y-3+y)/(y(3-y)) 4 = (2y-3)/(3y-y^2) 4(3y-y^2) = 2y-3 12y - 4y^2 = 2y -3 -4y^2 + 10y + 3 = 0 ------------>Since the 3 is positive, there are no positive roots to this equation, meaning that there are no value for which both x and y are positive.

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