Recent content by Fredh

Oh! Of course, it didn't occur to me it was y(0) I had to use, thank you very much!

I don't know what to do, can you tell me how does it fit?

Edit: trying to solve it again

Thanks! It was not a typo, anyway, I solved yp' = p + y²: dp/dy - p/y = y Integrating factor = 1/y p'/y - p/y² = 1 And that gives me y' = y(y+k) But that does not fit the initial conditon y'(0) = 1

Homework Statement y*y'' = (y')² + y²*y' y(0) = -1/2 , y'(0) = 1 Homework Equations The Attempt at a Solution y' = p y'' = p'p p(p'p) = p² + y²p Dividing both sides by p p'p = p + y² And then I'm stuck

Hey vela, thanks again, I came back to this problem today and solved it finnally, I was having trouble finding the recurrence formula a_{k+2}=ka_{k}-a_{k-2}/(k+1)(k+2)

$$2a_2 + (6a_3-a_1)x + \sum_{n=2}^\infty [(n+2)(n+1)a_{n+2} - na_n + a_{n-2}]x^n = 0$$ Ok, I must be doing something very wrong here, tried setting $$a_0 = 1 , a_1 = 0$$ but I can't really take anything usefull out of it Isn't $$a_0 = 2a_2 = 1$$ in this case? And after finding one...

Hey Vela, thanks! I got back to the problem this morning and found the expression you wrote. I'm going to finnish it after lunch and post results.

I didn't get it, so I should make a sum from 2 on and ignore the smaller terms? I'm really lost here, there are no similar examples in my material (where I have to make the k > n transformation more than once) yet all problems are like this

Homework Statement y'' - xy' + x²y = 0 Homework Equations y = Ʃ An*x^n (from 0 to infinity) y' = Ʃ n*An*x^n-1 (from 1 to infinity) y'' = Ʃ n*(n-1)*An*x^n-2 (from 2 to infinity) The Attempt at a Solution Ʃ n*(n-1)*An*x^n-2 (from 2 to infinity) - Ʃ n*An*x^n (from 1 to...