The differential equation is:

$\displaystyle y'xlnx + y(lnx + 1) = 1$

for x > 1

How can I simplify this?

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- Nov 27th 2010, 05:00 AMjenkkiHow can I solve this differential equation?
__The differential equation is__:

$\displaystyle y'xlnx + y(lnx + 1) = 1$

for x > 1

How can I simplify this? - Nov 27th 2010, 05:08 AMjenkki
Solved it; the left side can be rewritten as $\displaystyle [yxlnx]'$

- Nov 27th 2010, 05:10 AMchisigma
Setting $\displaystyle \ln x = t$ the DE becomes...

$\displaystyle \displaystyle t\ \frac{dy}{dt} + (1+t)\ y = 1$ (1)

... the solution of which is a 'standard job'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 27th 2010, 05:34 AMPandevil1990