y = ∫y'dx = ∫(2/√x - x)dx

y = 2∫x

^{-½}dx - ∫xdxy = 4√x - ½x

^{2}+ CSince (2,1) lies on the curve, y = 1 when x = 2.

So, 1 = 4√2 - 2 + C

C = 3 - 4√2

y = 4√x - ½x

^{2}+ 3 - 4√2Marcus A.

asked • 09/08/17Hi, can anybody help me with the following question? "Find the equation of the curve with the derivative of y with respect to x given as (2/√x)-x, passing through the point (2,1)"

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y = ∫y'dx = ∫(2/√x - x)dx

y = 2∫x^{-½}dx - ∫xdx

y = 4√x - ½x^{2} + C

Since (2,1) lies on the curve, y = 1 when x = 2.

So, 1 = 4√2 - 2 + C

C = 3 - 4√2

y = 4√x - ½x^{2} + 3 - 4√2

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Marcus A.

09/08/17