the last integral does not integrate :

$\displaystyle xy' - y=ln(y); x=(ln(y) + y)/y'; y' = p; x=(ln(y) + y)/p; $

$\displaystyle dx=-1/p^2 (ln(y) + y)dp + (1/yp + 1/p)dy; $

$\displaystyle (ln(y)/p^2 + y/p^2)dp=dy/yp; dp/p = dy/(yln(y)+y^2)$

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- Mar 31st 2010, 11:58 PMnoteilerwhats wrong with my solution
the last integral does not integrate :

$\displaystyle xy' - y=ln(y); x=(ln(y) + y)/y'; y' = p; x=(ln(y) + y)/p; $

$\displaystyle dx=-1/p^2 (ln(y) + y)dp + (1/yp + 1/p)dy; $

$\displaystyle (ln(y)/p^2 + y/p^2)dp=dy/yp; dp/p = dy/(yln(y)+y^2)$ - Apr 1st 2010, 02:45 AMmr fantastic
The solution involves an integral that cannot be found using a finite number of elementary functions: integrate 1/(Log[y] + y) - Wolfram|Alpha

Where has the DE come from? - Apr 1st 2010, 05:44 AMnoteiler
this DE was given to me by university tutor. and yes, i know that function doesn't integrate in elementary functions, but i thought that i am doing something wrong. by the way i have plenty DEs that i can't solve duo to integration for example in attachment. i think that i'm just doing something wrong. can someone help me.