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andriy [413]
3 years ago
7

– 36 – m = 8(4 + 2m)

Mathematics
2 answers:
valentina_108 [34]3 years ago
5 0

Answer:

m = -4

Step-by-step explanation:

- 36 - m = 8(4+2m)

- 36 - m = 32+16m (to remove m from the left side, we need to add m to both sides)

- 36 = 32+17m (subtract 32 from both sides)

- 68 = 17m (divide by 17 on both sides)

-4 = m

Marina86 [1]3 years ago
5 0

Answer:

-36-m=32+16m

-m-16m=32+36

-17m=68

m=68/-17

m=-4

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Xy''+2y'-xy by frobenius method
aalyn [17]
First note that x=0 is a regular singular point; in particular x=0 is a pole of order 1 for \dfrac2x.

We seek a solution of the form

y=\displaystyle\sum_{n\ge0}a_nx^{n+r}

where r is to be determined. Differentiating, we have

y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}
y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}

and substituting into the ODE gives

\displaystyle x\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-x\sum_{n\ge0}a_nx^{n+r}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge2}a_{n-2}x^{n+r-1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}\bigg((n+r)(n+r+1)a_n-a_{n-2}\bigg)x^{n+r-1}=0

The indicial polynomial, r(r+1), has roots at r=0 and r=-1. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.

When r=0, we have the recurrence

a_n=\dfrac{a_{n-2}}{(n+1)(n)}

valid for n\ge2. When n=2k, with k\in\{0,1,2,3,\ldots\}, we find

a_0=a_0
a_2=\dfrac{a_0}{3\cdot2}=\dfrac{a_0}{3!}
a_4=\dfrac{a_2}{5\cdot4}=\dfrac{a_0}{5!}
a_6=\dfrac{a_4}{7\cdot6}=\dfrac{a_0}{7!}

and so on, with a general pattern of

a_{n=2k}=\dfrac{a_0}{(2k+1)!}

Similarly, when n=2k+1 for k\in\{0,1,2,3,\ldots\}, we find

a_1=a_1
a_3=\dfrac{a_1}{4\cdot3}=\dfrac{2a_1}{4!}
a_5=\dfrac{a_3}{6\cdot5}=\dfrac{2a_1}{6!}
a_7=\dfrac{a_5}{8\cdot7}=\dfrac{2a_1}{8!}

and so on, with the general pattern

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

So the first indicial root admits the solution

y=\displaystyle a_0\sum_{k\ge0}\frac{x^{2k}}{(2k+1)!}+a_1\sum_{k\ge0}\frac{x^{2k+1}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}

which you can recognize as the power series for \dfrac{\sinh x}x and \dfrac{\cosh x}x.

To be more precise, the second series actually converges to \dfrac{\cosh x-1}x, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When r=-1, we may seek a second solution of the form

y=cy_1\ln x+x^{-1}\displaystyle\sum_{n\ge0}b_nx^n

where y_1=\dfrac{\sinh x+\cosh x-1}x. Substituting this into the ODE, you'll find that c=0, and so we're left with

y=x^{-1}\displaystyle\sum_{n\ge0}b_nx^n
y=\dfrac{b_0}x+b_1+b_2x+b_3x^2+\cdots

Expanding y_1, you'll see that all the terms x^n with n\ge0 in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form y_2=\dfrac1x. Adding this to y_1, we end up with just \dfrac{\sinh x+\cosh x}x.

This means the general solution for the ODE is

y=C_1\dfrac{\sinh x}x+C_2\dfrac{\cosh x}x
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3 years ago
You are saving for a spring break cruise for the end of march of the next year thag costs $600. You start at $50 in August. At t
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March is 8 months away from August and the target amount is $600.
August = $50
September = $50 + (0.2*50)=$60
October = $60 + (0.2*60)=$72
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December=$86.4 + (0.2*$86.4)=$103.68
January=$103.68+(0.2*103.68)= $124.416
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Total amount = $824.6

Yes, she will have money for the trip
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Answer:

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