x² y' + 2xy =	2 ln x
	x

⇒ y' +	2	y =	2 ln x
	x		x³

Comparing, we get P =	2	Q =	2 ln x
	x'		x³

y(I.F.) = ∫ Q I.F. dx + c

x² y = ∫	2 ln x	. x²dx + C
	x³

Solving above, find the value of C.
Putting x = 1, then find the value of y at x = e.

Which is y =	1
	e²

If x²		dy	+ 2xy =	2ln(x)	, and y(1) = 0, then what is y(e)?
		dx		x