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ziro4ka [17]
2 years ago
11

Help me please i need it rn

Mathematics
1 answer:
umka2103 [35]2 years ago
5 0

Step-by-step explanation:

Multiply 12600 by -3/4 to determine the amount earned.

*12600 can be put over 1.

<u>Multiply:</u>

\frac{12600}{1}  \times  \frac{ - 3}{4}  =  \frac{37800}{4}

<u>Simplify by dividing the fraction by 4:</u>

\frac{ - 37800 \div 4}{4 \div 4}  =  \frac{ - 9450}{1}

Mikasa has lost $9450.

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Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a
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I'll assume the ODE is actually

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Look for a series solution centered at x=2, with

y=\displaystyle\sum_{n\ge0}c_n(x-2)^n

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\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n

with c_0=y(2)=2 and c_1=y'(2)=0.

Substituting the series into the ODE gives

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0

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  • If n=2k for integers k\ge0, then

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k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}

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c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}

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So we have

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