The derivative of an implicit function is known and the implicit function is solved Given (y-2xy) DX + x2dy = 0, when x = 1, y = E; find y = f (x) 2 in x2dy is the square of X

A:
(y-2xy) DX + x ^ 2dy = 0, written as
x^2dy = y(2x-1)dx
That is dy / y = (2x-1) / x ^ 2 DX
The integral of both sides is as follows:
ln|y|=2ln|x|+1/x+C
Substituting x = 1, y = e, the solution is C = 0
So ln | y | = 2ln | x | + 1 / X
y=x^2*e^(1/x)

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