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3 years ago

How do you write out x <=-4

1 answer:

3 0

You can write it out multiple ways, such as x<=-4, x is (-infinity, -4], or {x | x<=4}. Even using a line, put -4 down, with a closed circle (including it means) and an arrow from there pointing left, for the included values to negative infinity.

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Find two linearly independent power series solutions about the point x0 = 0 of

aksik [14]

Assume a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting into the ODE, which appears to be

$(x^2-4)y''+3xy'+y=0,$

gives

$\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}x^{n+2}-4(n+2)(n+1)a_{n+2}x^n+3(n+1)a_{n+1}x^{n+1}+a_nx^n\bigg)=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^n-4\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n+3\sum_{n\ge1}na_nx^n+\sum_{n\g0}a_nx^n=0$

$(a_0-8a_2)+(4a_1-24a_3)x+\displaystyle\sum_{n\ge2}\bigg[(n+1)^2a_n-4(n+2)(n+1)a_{n+2}\bigg]x^n=0$

which gives the recurrence for the coefficients $a_n$ ,

$\begin{cases}a_0=a_0\\a_1=a_1\\4(n+2)a_{n+2}=(n+1)a_n&\text{for }n\ge0\end{cases}$

There's dependency between coefficients that are 2 indices apart, so we consider 2 cases.

If $n=2k$ , where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

$k=1\implies n=2\implies a_2=\dfrac1{4\cdot2}a_0=\dfrac2{4\cdot2^2}a_0=\dfrac{2!}{2^4}a_0$

$k=2\implies n=4\implies a_4=\dfrac3{4\cdot4}a_2=\dfrac3{4^2\cdot4\cdot2}a_0=\dfrac{4!}{2^8(2!)^2}a_0$

$k=3\implies n=6\implies a_6=\dfrac5{4\cdot6}a_4=\dfrac{5\cdot3}{4^3\cdot6\cdot4\cdot2}a_0=\dfrac{6!}{2^{12}(3!)^2}a_0$

and so on, with the general pattern

$a_{2k}=\dfrac{(2k)!}{2^{4k}(k!)^2}a_0$

If $n=2k+1$ , then

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

$k=1\implies n=3\implies a_3=\dfrac2{4\cdot3}a_1=\dfrac{2^2}{2^2\cdot3\cdot2}a_1=\dfrac1{(3!)^2}a_1$

$k=2\implies n=5\implies a_5=\dfrac4{4\cdot5}a_3=\dfrac{4\cdot2}{4^2\cdot5\cdot3}a_1=\dfrac{(2!)^2}{5!}a_1$

$k=3\implies n=7\implies a_7=\dfrac6{4\cdot7}a_5=\dfrac{6\cdot4\cdot2}{4^3\cdot7\cdot5\cdot3}a_1=\dfrac{(3!)^2}{7!}a_1$

and so on, with

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

Then the two independent solutions to the ODE are

$\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{(2k)!}{2^{4k}(k!)^2}x^{2k}}$

and

$\boxed{y_2(x)=\displaystyle a_1\sum_{k\ge0}\frac{(k!)^2}{(2k+1)!}x^{2k+1}}$

By the ratio test, both series converge for $|x|$ , which also can be deduced from the fact that $x=\pm2$ are singular points for this ODE.

6 0

3 years ago

2 (4x-2)- 5x=-18<br> HELP!!

Arturiano [62]

I think it’s x= -14/3

8 0

3 years ago

Read 2 more answers

24 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

baherus [9]

60 * 100 / 65 = 92.31

100% - 92.31% = 7.69%

Answer: Decrease 7.69%

3 0

4 years ago

Read 2 more answers

please answer both questions correctly for 7 points + i will also mark you brainliest but only if you answer both correctly.

yKpoI14uk [10]

Question 12

Given

Point (-1, 5)

Slope m = -1

The slope-intercept form of the line equation

$y = mx+b$

where m is the slope and b is the y-intercept

In our case:

Point (-1, 5)

m = -1

substituting m = -1, and the point (-1, 5)

y = mx+b

5 = (-1)(-1) + b

5 = 1+b

b = 5-1

b = 4

Therefore, the value of y-intercept b = 4

now substituting m = -1 and b = 4 in the slope-intercept form of the line equation

y = mx+b

y = (-1)x + 4

y = -x+4

Thus, the slope-intercept of the line equation containing the point (-1, 5) and slope = -1 is:

y = -x+4

Question 13

Given

Point (3, 3)

Slope m = 2

The slope-intercept form of the line equation

$y = mx+b$

where m is the slope and b is the y-intercept

In our case:

Point (3, 3)

m = 2

substituting m = 2, and the point (3, 3)

y = mx+b

3 = (2)(3) + b

3 = 6+b

b = 3-6

b = -3

Therefore, the value of y-intercept b = -3

now substituting m = 2 and b = -3 in the slope-intercept form of the line equation

y = mx+b

y = (2)x + (-3)

y = 2x-3

Thus, the slope-intercept of the line equation containing the point (3, 3) and slope = -3 is:

y = 2x-3

Therefore, option A is true.

3 0

3 years ago

5\6 of 45kg plz can u tell me ans

LekaFEV [45]

37.5
You are very welcome

8 0

3 years ago