Simple first-order series solution

  • #1
th3chemist
66
0

Homework Statement



The question is y' - xy = 0
I have to solve it using series solutions

Homework Equations





The Attempt at a Solution



I use y = Ʃ from 0 to infinity a_n x^n and took the derivative. I plugged it into the equation

I got the recurrence relation to be a1 = 0 and (k+1)a_(k+1) - a_(k-1) = 0
 
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  • #2
th3chemist said:

Homework Statement



The question is y' - xy = 0
I have to solve it using series solutions

Homework Equations





The Attempt at a Solution



I use y = Ʃ from 0 to infinity a_n x^n and took the derivative. I plugged it into the equation

I got the recurrence relation to be a1 = 0 and (k+1)a_(k+1) - a_(k-1) = 0

Assuming that your work is correct (can't tell because you don't show it), then consider a0 to be arbitrary, and find a2 in terms of a0. a3 will be zero because a1 is. Continue finding a4, a6, and so on, in terms of a0.
 
  • #3
That is, if [itex]a_0[/itex] is the inital value, taking k= 1, [itex]2a_2= a_0[/itex] so [itex]a_2= a_0/2[/itex]. Then taking k= 2 [itex]3a_3= a_1= 0[/itex] so [itex]a_3= 0[/itex]. Taking k= 3, [itex]4a_4= a_2= a_0/2[/itex] so [itex]a_4= a_0/8[/itex]. Taking k= 4, [itex]5a_5= a_3= 0[/itex] so [itex]a_5= 0[/itex]. Taking k= 5, [itex]6a_6= a_4= a_0/8[/itex] so [itex]a_6= a_0/48[/itex], etc.

It should be obvious that [itex]a_n= 0[/itex] for all odd n (and you should be able to prove it by induction on n) while for n even we get [itex]a_0[/itex] over 2, 8, 48, ... Continue for a few more terms to see if you can find a formula for those denominators. It shouldn't surprise you if it involves a factorial.
 
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