Find dy/dx for the relation x^2y + xlnx = 3y, x > 0
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Find dy/dx for the relation x^2y + xlnx = 3y, x > 0

[From: ] [author: ] [Date: 12-05-01] [Hit: ]
(x^2 - 3)(dy/dx) = -2xy - ln(x) - 1.Finally,dy/dx = [-2xy - ln(x) - 1]/(x^2 - 3).I hope this helps!......
could you provide a solution, please and thank you

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Differentiating both sides with respect to x yields:
(d/dx)[x^2*y + x*ln(x)] = (d/dx)(3y) ==> 2xy + x^2(dy/dx) + ln(x) + 1 = 3(dy/dx).

Note that the product rule was applied twice: once on x^2*y and once on x*ln(x). Also, notice that y is a function of x, so the derivative of y is dy/dx.

Bringing all terms that involve dy/dx to one side and the rest to the other yields:
x^2(dy/dx) - 3(dy/dx)= -2xy - ln(x) - 1.

Factoring out dy/dx from the left side:
(x^2 - 3)(dy/dx) = -2xy - ln(x) - 1.

Finally, dividing both sides by x^2 - 3 gives:
dy/dx = [-2xy - ln(x) - 1]/(x^2 - 3).

I hope this helps!
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