The differential equation is: $\displaystyle y'xlnx + y(lnx + 1) = 1$ for x > 1 How can I simplify this?
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Solved it; the left side can be rewritten as $\displaystyle [yxlnx]'$
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 $\displaystyle \chi$ $\displaystyle \sigma$
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