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1 year ago

Less than, greater than, or equal to

Mathematics

1 answer:

liq [111]1 year ago

4 0

Answer:

equal to

Step-by-step explanation:

when two negatives are multiplied they cancel out leaving the positive 5. This means they are equal. 5=5

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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

$y=C_1\dfrac{\sinh x}x+C_2\dfrac{\cosh x}x$