The answers would be any divisor of 36 that fits into this domain 1 < x < 10.
This would be 2, 3, 4, 6, 9
Answer:496
Step-by-step explanation:
We want to find a solution such that
![y=\displaystyle\sum_{n\ge0}a_nx^n](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7Da_nx%5En)
![y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n](https://tex.z-dn.net/?f=y%27%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%28n%2B1%29a_%7Bn%2B1%7Dx%5En)
![y''=\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n](https://tex.z-dn.net/?f=y%27%27%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%28n%2B1%29%28n%2B2%29a_%7Bn%2B2%7Dx%5En)
With these conditions,
and
.
Substituting the series into the ODE gives
![\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n+3\sum_{n\ge0}(n+1)a_{n+1}x^n=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%28n%2B1%29%28n%2B2%29a_%7Bn%2B2%7Dx%5En%2B3%5Csum_%7Bn%5Cge0%7D%28n%2B1%29a_%7Bn%2B1%7Dx%5En%3D0)
![\displaystyle\sum_{n\ge0}\bigg((n+1)(n+2)a_{n+2}+3(n+1)a_{n+1}\bigg)x^n=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%5Cbigg%28%28n%2B1%29%28n%2B2%29a_%7Bn%2B2%7D%2B3%28n%2B1%29a_%7Bn%2B1%7D%5Cbigg%29x%5En%3D0)
so that the coefficients of
satisfy
![(n+1)(n+2)a_{n+2}+3(n+1)a_{n+1}=0](https://tex.z-dn.net/?f=%28n%2B1%29%28n%2B2%29a_%7Bn%2B2%7D%2B3%28n%2B1%29a_%7Bn%2B1%7D%3D0)
for
, or
![(n-1)na_n+3(n-1)a_{n-1}=0\implies a_n=-\dfrac3na_{n-1}](https://tex.z-dn.net/?f=%28n-1%29na_n%2B3%28n-1%29a_%7Bn-1%7D%3D0%5Cimplies%20a_n%3D-%5Cdfrac3na_%7Bn-1%7D)
for
. Notice that this implies a dependency of all
beyond
on
, while
takes on whatever initial value
is given. In particular,
![a_2=-\dfrac32a_1](https://tex.z-dn.net/?f=a_2%3D-%5Cdfrac32a_1)
![a_3=-\dfrac33a_2=\dfrac{3^2}{3\cdot2}a_1=\dfrac{3^2}{3!}a_1](https://tex.z-dn.net/?f=a_3%3D-%5Cdfrac33a_2%3D%5Cdfrac%7B3%5E2%7D%7B3%5Ccdot2%7Da_1%3D%5Cdfrac%7B3%5E2%7D%7B3%21%7Da_1)
![a_4=-\dfrac34a_3=-\dfrac{3^3}{4\cdot3\cdot2}a_1=-\dfrac{3^3}{4!}a_1](https://tex.z-dn.net/?f=a_4%3D-%5Cdfrac34a_3%3D-%5Cdfrac%7B3%5E3%7D%7B4%5Ccdot3%5Ccdot2%7Da_1%3D-%5Cdfrac%7B3%5E3%7D%7B4%21%7Da_1)
and so on up to
![a_n=\dfrac{(-3)^{n-1}}{n!}a_1](https://tex.z-dn.net/?f=a_n%3D%5Cdfrac%7B%28-3%29%5E%7Bn-1%7D%7D%7Bn%21%7Da_1)
So we can extract two fundamental solutions
such that
, where
![y_1=a_0](https://tex.z-dn.net/?f=y_1%3Da_0)
![y_2=\displaystyle-3a_1\sum_{n\ge1}\frac{(-3x)^n}{n!}](https://tex.z-dn.net/?f=y_2%3D%5Cdisplaystyle-3a_1%5Csum_%7Bn%5Cge1%7D%5Cfrac%7B%28-3x%29%5En%7D%7Bn%21%7D)
Recall that
![e^x=\displaystyle\sum_{n\ge0}\frac{x^n}{n!}](https://tex.z-dn.net/?f=e%5Ex%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%5Cfrac%7Bx%5En%7D%7Bn%21%7D)
which tells us
![y_2=-3a_1(e^{-3x}-1)=-3a_1e^{-3x}+3a_1](https://tex.z-dn.net/?f=y_2%3D-3a_1%28e%5E%7B-3x%7D-1%29%3D-3a_1e%5E%7B-3x%7D%2B3a_1)
but
is a constant solution and already accounts for the constant term in
, and
can be reduced to a simpler constant
, leaving us with
![y_2=a_1e^{-3x}](https://tex.z-dn.net/?f=y_2%3Da_1e%5E%7B-3x%7D)
The Wronskian is
![W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=\begin{vmatrix}a_0&a_1e^{-3x}\\0&-3a_1e^{-3x}\end{vmatrix}=-3a_0a_1e^{-3x}](https://tex.z-dn.net/?f=W%28y_1%2Cy_2%29%3D%5Cbegin%7Bvmatrix%7Dy_1%26y_2%5C%5C%7By_1%7D%27%26%7By_2%7D%27%5Cend%7Bvmatrix%7D%3D%5Cbegin%7Bvmatrix%7Da_0%26a_1e%5E%7B-3x%7D%5C%5C0%26-3a_1e%5E%7B-3x%7D%5Cend%7Bvmatrix%7D%3D-3a_0a_1e%5E%7B-3x%7D)
![\implies W(y_1,y_2)(0)=-3a_0a_1](https://tex.z-dn.net/?f=%5Cimplies%20W%28y_1%2Cy_2%29%280%29%3D-3a_0a_1)
so the two solutions are indeed independent as long as neither initial value is 0.