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ValentinkaMS [17]
3 years ago
13

A hospital discards 1000 patient gowns every 3 months how many gallons will they need to replace the discarded ones over the nex

t year
Mathematics
1 answer:
vampirchik [111]3 years ago
4 0
Keeping in mind that, well, first off, they're not getting any gallons of gowns btw, anyway.... how many gowns will they need for a year?   well, depends on how many they discarded in a year.  Now, a year has 12 months, and we know they discard 1000 in 3 months, thus

\bf \begin{array}{ccll}
gowns&months\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
1000&3\\
g&12
\end{array}\implies \cfrac{1000}{g}=\cfrac{3}{12}\implies \cfrac{1000}{g}=\cfrac{1}{4}\implies 4000=g
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Let

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots

Differentiating twice gives

\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots

\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n

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\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0

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\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

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k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}

k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}

It should be easy enough to see that

a_{n=2k} = \dfrac{a_0}{(2k)!}

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

k = 0 \implies n=1 \implies a_1 = a_1

k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}

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k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}

so that

a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}

So, the overall series solution is

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\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}

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<h2>Answer: </h2>

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