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Y"-3xy=0 given ordinary point x=0. power series

User Xmak
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\displaystyle y=\sum_(n\ge0)a_nx^n

\implies\displaystyle y''=\sum_(n\ge2)n(n-1)a_nx^(n-2)


y''-3xy=0

\displaystyle\sum_(n\ge2)n(n-1)a_nx^(n-2)-3\sum_(n\ge0)a_nx^(n+1)=0

\displaystyle2a_2+\sum_(n\ge3)n(n-1)a_nx^(n-2)-3\sum_(n\ge0)a_nx^(n+1)=0

\displaystyle2a_2+\sum_(n\ge3)n(n-1)a_nx^(n-2)-3\sum_(n\ge3)a_(n-3)x^(n-2)=0

\displaystyle2a_2+\sum_(n\ge3)\bigg(n(n-1)a_n-3a_(n-3)\bigg)x^(n-2)=0

This generates the recurrence relation


\begin{cases}a_0=a_0\\a_1=a_1\\2a_2=0\\n(n-1)a_n-3a_(n-3)=0&\text{for }n\ge3\end{cases}

Because you have


n(n-1)a_n-3a_(n-3)=0\implies a_n=\frac3{n(n-1)}a_(n-3)

it follows that
a_2=0\implies a_5=a_8=a_(11)=\cdots=a_(n=3k-1)=0 for all
k\ge1.

For
n=1,4,7,10,\ldots, you have


a_1=a_1

a_4=\frac3{4*3}a_1=(3*2)/(4!)a_1

a_7=\frac3{7*6}a_4=(3*5)/(7*6*5)=(3^2*5*2)/(7!)

a_(10)=\frac3{10*9}a_7=(3*8)/(10*9*8)a_7=(3^3*8*5*2)/(10!)a_1

so that, in general, for
n=3k-2,
k\ge1, you have


a_(n=3k-2)=(3^(k-1)\displaystyle\prod_(\ell=1)^(k-1)(3\ell-1))/((3k-2)!)a_1

Now, for
n=0,3,6,9,\ldots, you have


a_0=a_0

a_3=\frac3{3*2}a_0=\frac3{3!}a_0

a_6=\frac3{6*5}a_3=(3*4)/(6*5*4)a_3=(3^2*4)/(6!)a_0

a_9=\frac3{9*8}a_6=(3*7)/(9*8*7)a_6=(3^3*7*4)/(9!)a_0

and so on, with a general pattern for
n=3k,
k\ge0, of


a_(n=3k)=(3^k\displaystyle\prod_(\ell=1)^k(3\ell-2))/((3k)!)a_0

Putting everything together, we arrive at the solution


y=\displaystyle\sum_(n\ge0)a_nx^n

y=a_0\underbrace{\displaystyle\sum_(k\ge0)(3^k\displaystyle\prod_(\ell=1)^k(3\ell-2))/((3k)!)x^(3k)}_(n=0,3,6,9,\ldots)+a_1\underbrace{\displaystyle\sum_(k\ge1)(3^(k-1)\displaystyle\prod_(\ell=1)^(k-1)(3\ell-1))/((3k-2)!)x^(3k-2)}_(n=1,4,7,10,\ldots)

To show this solution is sufficient, I've attached is a plot of the solution taking
y(0)=a_0=1 and
y'(0)=a_1=0, with
n=6. (I was hoping to be able to attach an animation that shows the series solution (orange) converging rapidly to the exact solution (blue), but no such luck.)
Y"-3xy=0 given ordinary point x=0. power series-example-1
User Vitor Villar
by
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