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chubhunter [2.5K]
3 years ago
8

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

Mathematics
1 answer:
liberstina [14]3 years ago
6 0
\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=\dfrac3{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=\dfrac3{4\times3}a_1=\dfrac{3\times2}{4!}a_1
a_7=\dfrac3{7\times6}a_4=\dfrac{3\times5}{7\times6\times5}=\dfrac{3^2\times5\times2}{7!}
a_{10}=\dfrac3{10\times9}a_7=\dfrac{3\times8}{10\times9\times8}a_7=\dfrac{3^3\times8\times5\times2}{10!}a_1

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

a_{n=3k-2}=\dfrac{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=\dfrac3{3\times2}a_0=\dfrac3{3!}a_0
a_6=\dfrac3{6\times5}a_3=\dfrac{3\times4}{6\times5\times4}a_3=\dfrac{3^2\times4}{6!}a_0
a_9=\dfrac3{9\times8}a_6=\dfrac{3\times7}{9\times8\times7}a_6=\dfrac{3^3\times7\times4}{9!}a_0

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

a_{n=3k}=\dfrac{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}\frac{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}\frac{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.)

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Enzo and Beatrice are playing games at their local arcade incredibly ends wins five tickets from every game in the trees wins 11
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Step-by-step explanation:

In order to find the minimum number of games that Enzo played we need to find the least common multiple of tickets where both Enzo and Beatrice are equal. To do this we simply find all the multiples of the number of games each could have played by the number of tickets per game until we find one in common.

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5*22 = 110               11*10 = 110

Finally, we can see that the minimum amount of games that Enzo needs to play in order for both Enzo and Beatrice to have the same amount of tickets is 22 games.

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