Simple first-order series solution

[th3chemist]

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

 

[Mark44]

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.

 

[HallsofIvy]

That is, if $a_0$ is the inital value, taking k= 1, $2a_2= a_0$ so $a_2= a_0/2$. Then taking k= 2 $3a_3= a_1= 0$ so $a_3= 0$. Taking k= 3, $4a_4= a_2= a_0/2$ so $a_4= a_0/8$. Taking k= 4, $5a_5= a_3= 0$ so $a_5= 0$. Taking k= 5, $6a_6= a_4= a_0/8$ so $a_6= a_0/48$, etc.

It should be obvious that $a_n= 0$ for all odd n (and you should be able to prove it by induction on n) while for n even we get $a_0$ 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.