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## Homework Statement

Find the general solution of

y' + (2/x)y = 3/(x^2)

## The Attempt at a Solution

xy' + 2y = 3/x

d/dx (x * 2y) = 3/x

integrating both sides (using product rule for LHS) I end up with

y= (3lnx + C)/2x

Then I am supposed to find the solution for which y(2) = -1 so I subbed that into the above and got c = -4/3 - ln2

So then I can say that y = 3lnx/(2x) - 4/3 -ln2

Hope thats right!

Thanks guys!