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This is my solution so far : http://imgur.com/a/WuDww

I turned the equation into bernolli dif. eq. then to first degree linear dif eq. but im stuck after that. What to do afterwards ?

2 Answers 2

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hint: $(f.g)'=f'g+g'f$ $$x^2u'+2xu=xlnx\\(x^2u)'=xlnx$$then apply integral to both sides

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    You (and the OP) missed a minus sign in front of the first term; this will change the integrating factor. – 2017-01-23
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$-u' + \frac 2x u = \frac {\ln x}{x}$

You need to flip the sign on $u'$ before looking for an integrating factor.

$u' - \frac 2x u = -\frac {\ln x}{x}$

Integrating factor $e^{\int -\frac 2x} = e^{-2\ln x} = x^{-2}$

$x^{-2} u' - 2x^{-3} u = -x^{-3}\ln x\\ x^{-2} u = \int -x^{-3}\ln x\; dx$

And you should be able to get home from here.

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    thanks for the solution but can you fix the second step's text ? – 2017-01-23