Solve y''= x^3/y I tried by using let u=y' then I get u'=x^3/y .. now something tells me this is not going to be an easy ride home.
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Lets write the second order DE as... (1) ... and suppose that the solution is analytic in so that is... (2a) (2b) (2c) The are found inserting the (2a) and (2c) in (1), once we know the 'initial conditions' and ... ... and so one... Kind regards
Why assume x=0? Why not just start with u=y' ?
Originally Posted by Khonics89 Why assume x=0? Why not just start with u=y' ? Because then you get a single equation with the two unknown functions, u and y. And he did NOT "assume x= 0", he assumed the solution was analytic at x= 0.
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