Question

Practicing LaTeX.

Answer 1

Aiight i'm here

Answer 2

\[\iiint_\mathbb{C}\]

Answer 3

\begin{array}l\color{#FF0000}{\text{LaTeX}}\color{#FF0000}{\text{ }}\color{#FF7F00}{\text{is}}\color{#FF7F00}{\text{ }}\color{#FFE600}{\text{awesome.}}\color{#FFE600}{\text{ }}\color{#00FF00}{\text{I}}\color{#00FF00}{\text{ }}\color{#0000FF}{\text{can}}\color{#0000FF}{\text{ }}\color{#6600FF}{\text{type}}\color{#6600FF}{\text{ }}\color{#8B00FF}{\text{like}}\color{#8B00FF}{\text{ }}\color{#FF0000}{\text{this!}}\color{#FF0000}{\text{ }}\end{array}

Answer 4

\begin{array}l\color{#FF0000}{\text{O}}\color{#FF7F00}{\text{r}}\color{#FFE600}{\text{,}}\color{#00FF00}{\text{ }}\color{#00FF00}{\text{I}}\color{#0000FF}{\text{ }}\color{#0000FF}{\text{c}}\color{#6600FF}{\text{a}}\color{#8B00FF}{\text{n}}\color{#FF0000}{\text{ }}\color{#FF0000}{\text{t}}\color{#FF7F00}{\text{y}}\color{#FFE600}{\text{p}}\color{#00FF00}{\text{e}}\color{#0000FF}{\text{ }}\color{#0000FF}{\text{l}}\color{#6600FF}{\text{i}}\color{#8B00FF}{\text{k}}\color{#FF0000}{\text{e}}\color{#FF7F00}{\text{ }}\color{#FF7F00}{\text{t}}\color{#FFE600}{\text{h}}\color{#00FF00}{\text{i}}\color{#0000FF}{\text{s}}\color{#6600FF}{\text{!}}\end{array}

Answer 5

Ask away.

Answer 6

\begin{array}l\color{#FF0000}{\text{I}}\color{#FF7F00}{\text{ }}\color{#FF7F00}{\text{i}}\color{#FFE600}{\text{m}}\color{#00FF00}{\text{p}}\color{#0000FF}{\text{r}}\color{#6600FF}{\text{e}}\color{#8B00FF}{\text{s}}\color{#FF0000}{\text{s}}\color{#FF7F00}{\text{ }}\color{#FF7F00}{\text{a}}\color{#FFE600}{\text{l}}\color{#00FF00}{\text{l}}\color{#0000FF}{\text{ }}\color{#0000FF}{\text{t}}\color{#6600FF}{\text{h}}\color{#8B00FF}{\text{e}}\color{#FF0000}{\text{ }}\color{#FF0000}{\text{g}}\color{#FF7F00}{\text{i}}\color{#FFE600}{\text{r}}\color{#00FF00}{\text{l}}\color{#0000FF}{\text{s}}\color{#6600FF}{\text{ }}\color{#6600FF}{\text{b}}\color{#8B00FF}{\text{y}}\color{#FF0000}{\text{ }}\color{#FF0000}{\text{t}}\color{#FF7F00}{\text{y}}\color{#FFE600}{\text{p}}\color{#00FF00}{\text{i}}\color{#0000FF}{\text{n}}\color{#6600FF}{\text{g}}\color{#8B00FF}{\text{ }}\color{#FF0000}{\text{l}}\color{#FF7F00}{\text{i}}\color{#FFE600}{\text{k}}\color{#00FF00}{\text{e}}\color{#0000FF}{\text{ }}\color{#0000FF}{\text{t}}\color{#6600FF}{\text{h}}\color{#8B00FF}{\text{i}}\color{#FF0000}{\text{s}}\color{#FF7F00}{\text{ }}\color{#FF7F00}{\text{:}}\color{#FFE600}{\text{-}}\color{#00FF00}{\text{D}}\end{array}

Answer 7

Okay. Go away now.

Answer 8

y = Ax^3 + B/x^4 - x^2 /3 then x^2y''+2xy'-12y=2x^2

find the particular solution if y(1) = 2/3 and y(-1)=14/3

Answer 9

So you need to differentiate and plug that stuff in. Then plug in 1 for x and -1 for x. Then solve for A and B. You'll have two linear equations and two unknowns.

Answer 10

Then use elimination to solve for the 2 unknowns?

Answer 11

Okay, first question, how would i know what variable to plug it into if i had like a z or something in there (or is that something i'll ever have to worry about)

Answer 12

Because there are x's with all the y and y',y'' etc. terms. And your solution is in terms of x. So once you plug that in, you have an equation ONLY in terms of x.

Answer 13

And if there are more than 1 variable, they would have to specify.

Answer 14

gotcha

Answer 15

Jesus this problem is going to suck wingspan... lol

Answer 16

Yeahhhh They changed it to censor cussing... Its dumb.

Answer 17

LOLOLOLol

Answer 18

What next?

Answer 19

well lemme try and figure this one out real quick

Answer 20

i'm a little bit confused though, like you can plug the one into the equation but where does this = 2/3 because i can't see how its ever going to do that

Answer 21

What? Haha. You should have a linear equation in terms of A and B, then another in terms of A and B. Solve them how you see fit.

Answer 22

Type out what you get once you differentiate and plug everything in

Answer 23

Everything but the 1/-1

Answer 24

x^2(6Ax+20B/x^6-2/3)+2x(3Ax^2-4B/x^5-2x/3)-12(Ax^3+B/x^4-x^2/3) = 2x^2

Answer 25

Hrm... so there is no particular solution?

Answer 26

But what does the 2/3 mean where it has y(1) = 2/3

Answer 27

Ohh okay, that's what i was thinking

Answer 28

Alright
STOP

Answer 29

Yes?

Answer 30

I just told you the wrong thing. So your y equation is what?

Answer 31

you mean what i differentiated? y = Ax^3 + B/x^4 - x^2 / 3 ?

Answer 32

Yes. You have y(1) and y(-1) Plug that into THAT.
So you have:
\[\frac{2}{3}=A+B-\frac{1}{3} \implies 1=A+B \implies 1-B=A\]
Plug in -1.
\[\frac{14}{3}=-A+B-\frac{1}{3} \implies 5=-A+B \implies B=3; A=-2\]

Answer 33

ah... haha... well then... what do ya know

Answer 34

So i gotta erase all this garbage haha... so i for real don't need to bother at all with that other equation?

Answer 35

Nah.

Answer 36

Well that makes this a flutter load easier than I thought it was goin to be

Answer 37

Dude... censored words... no fun

Answer 38

I know :/

Answer 39

Hey one last question on this one, how exactly did you figure out what A and B was?

Answer 40

Cause i've got written down A = 1 - B and -A+B = 5

so what do I do now?

Answer 41

Solve for them.
If you have:
y=3x+2
y=3x-5
You can solve for x and y right? Its the same thing.

Answer 42

\[-(1-B)+B=5 \implies -1+2B=5 \implies 2B=6 \implies B=3; A=1-B\] \[\implies A=1-3=-2\]

Answer 43

Lol, okay... cool, I got ya