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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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Novay_Z [31]

He can give at most 2 adult haircuts with the remaining time

<h3>How many adult haircuts at most can he give with the remaining time? </h3>

The inequality is given as:

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Also, we have

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Substitute C = 5 in 0.75C + 1.25A <= 7

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<u>Complete question</u>

Horace is a professional hair stylist. Let C represent the number of child haircuts and A represent the number of adult haircuts that Horace can give within 7 hours. 0.75C + 1.25A <= 7

Horace gave 5 child haircuts.

How many adult haircuts at most can he give with the remaining time?

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Please help me with all of them or at least 2 of them? Thank you so much.
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Step-by-step explanation:

The zeroes of f are the points where the graph f crosses the x-axis, which occur at x = -2, x = 0, x = 2.

Have a great day :)

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