uhhh
#help-10
glacial obsidian
What is it like?
dense tinsel
instead of there being a c and a b
there is x1 and x2
but for all intents and purposes it's the same thing
glacial obsidian
there is x1 and x2
Oh so it is the same, just with different letters
dense tinsel
yeah
so
lemme plug it in real qucik
is there a command here in discord to zoom in
into the image like top right corner
glacial obsidian
I don't know
warm shaleBOT
GMDennis
dense tinsel
what do you think?
glacial obsidian
That is correct
dense tinsel
lets goooooo
anyway that's the end of my math homework
have a good day (or night idk its 8:18 pm for me)
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
devout lichen
Hey so this says 1580.9 million does that mean 1,580,900,000?
obtuse pebbleBOT
Hey so this says 1580.9 million does that mean 1,580,900,000?
latent walrus
most likely
devout lichen
I'm looking at a city budet if that helps
latent walrus
then yeah, they do tend to write things as millions, what you said holds
versed nebula
I'm looking at a city budet if that helps
Yes it means 1,580,900,000
versed nebula
".close"
latent walrus
'.close'
devout lichen
.close
kind hollow
hi
static marten
what about it?
kind hollow
what about it?
wrong problem mb
unique bear
obtuse pebbleBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
obtuse pebbleBOT
@kind hollow Has your question been resolved?
uncut obsidian
i've been trying to find the radius of this circle for a full hour, all the sides of the square are equal to 10
obtuse pebbleBOT
i've been trying to find the radius of this circle for a full hour, all the side...
uncut obsidian
anyone know how to do this?
wooden cipher
mark the center of the circle
draw the radii from the center to the two sides it is tangent to
show me when youre done
uncut obsidian
wooden cipher
ok cool, lets call the center O
and the radius of the circle is r
what is OA?
(in terms of r)
uncut obsidian
oh shit finally
i see it
sqrt(2)*r = the hypotheneus
then *2
for a0
then
4sqrt(2)r is equal to ac
diagonal
right?
or is that too optimistic
wooden cipher
not quite
uncut obsidian
ah
wooden cipher
OA is (sqrt2)r yes?
wooden cipher
ok, what is OC?
wooden cipher
no, look at the circle
wooden cipher
yeah there you go
uncut obsidian
thank you bro that took me so long
wooden cipher
youre welcome
alpine bison
timid silo
I'm not sure how to go about this
obtuse pebbleBOT
I'm not sure how to go about this
timid silo
We are supposed to solve it through FTC
timid silo
can you represent the absolute value function in piecewise form?
chrome crypt
is at x = 5/2
and so we can then integrate for when the area is less than 0 using that sub function to the vertex
and then integrate from the vertex to the other point
alpine bison
so the integral is $2\int_{5/2}^5 2x-5$
warm shaleBOT
everg
alpine bison
chrome crypt
| 2x - 5 | = { (2x - 5), if x >= 5/2
-(2x - 5), if x < 5/2
You can also solve it this way.
warm shaleBOT
timid silo
,w 2.5(5)(1/2)
warm shaleBOT
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
Give a binary operation on the set of positive integers which makes it a group.
timid silo
Can't really think of anything
spice citrus
I might be overcomplicating this, but you can choose any group with the same cardinality as the positive integers and then use its operation
because there will be a bijection between the integers and that group
obtuse pebbleBOT
@timid silo Has your question been resolved?
timid silo
I might be overcomplicating this, but you can choose any group with the same car...
could you give an example
e.g. the group (Z, +)
drifting wraith
map it to all integers
and do addition
1 2 3 4 5 6
0 1 −1 2 −2 3
should work idk?
yeah, no doubt
timid silo
obtuse pebbleBOT
timid silo
im stuck trying to integrate this
worn yoke
power reduction formulas https://math.libretexts.org/Courses/Borough_of_Manhattan_Community_College/MAT_206_Precalculus/7%3A_Trigonometric_Identities_and_Equations/7.3%3A_Double-Angle%2C_Half-Angle%2C_and_Reduction_Formulas
obtuse pebbleBOT
@timid silo Has your question been resolved?
worn yoke
you can apply it inside the square, then simplify, and apply again
close, but you should be careful of where your exponents go
rotund vector
Could someone help me out with this
obtuse pebbleBOT
Could someone help me out with this
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
burnt thunder
obtuse pebbleBOT
burnt thunder
,rotate 270
warm shaleBOT
obtuse pebbleBOT
,rotate 270
timid silo
how do i do 2
Kinda like this
timid silo
the parabola one
Oh
That just means when the derivative equals 0
So find the derivative
Then set it equal to 0
timid silo
so i just sub x for 0?
No
Set it equal to 0
Given the function x^2
2x is the derivative
2x=0
x=0 is your answer for when it is a horizontal tangent
timid silo
Oh wow
Hm
Find the vertex then @burnt thunder
The vertex is the middle point of the parabola
@burnt thunder
This is how you find the x value of the vertex
To find the y value you can plug that value right back in to the original function
@burnt thunder
You there?
obtuse pebbleBOT
@burnt thunder Has your question been resolved?
frigid holly
"What does 'antisymmetric' mean in the context of sets? Considering the relation R = {(A, B), (B, C), (D, E)}, I know that its antisymmetric but I cant understand it. I understand reflexive, transitive and symmetric
obtuse pebbleBOT
"What does 'antisymmetric' mean in the context of sets? Considering the relation...
latent walrus
asymmetric doesnt allow for reflexivity, antisymmetric does
ie with antisymmetric, if (a,b) in R and (b,a) are in R then a=b
asymmetric is (a,b) in R means (b,a) cant be in R, even if a=b
frigid holly
oh okay thx
sonic turtle
can someone tell me why the second is wrong asap
obtuse pebbleBOT
can someone tell me why the second is wrong asap
daring sorrel
Reduced means zeros above the pivots aswell
sonic turtle
OHHH
Shoot I didn't know that, so the entries above and below the leading 1's have to be zero right?
so vertical columns
?
daring sorrel
Yeah
sonic turtle
is this correct
daring sorrel
Yeah
brisk hamlet
obtuse pebbleBOT
@sonic turtle Has your question been resolved?
dark vector
Seeking help with this. here's what i have so far:
Suppose the existence of $a, b, c, d \in \mathbb{N}$ such that Lehman's equation holds.
Let $S_{a}$ be the set of possible values for $a$. Assume for contradiction's sake that $S_{a}$ is nonempty. Then by the Well Ordering Principle, $S_{a}$ must have some smalelst element $a_{0}.$
for context this was on a worksheet based on the WOP, which is why i've broken it out.
obtuse pebbleBOT
Seeking help with this. here's what i have so far:
Suppose the existence of $a,...
warm shaleBOT
Out Of Nosh
obtuse pebbleBOT
@dark vector Has your question been resolved?
dark vector
<@&286206848099549185>
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
little stirrup
how should I approach this I'm stuck
obtuse pebbleBOT
how should I approach this I'm stuck
little stirrup
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
thorn summit
how is the sum to product rule derived? I get the product to sum rule, but I can't understand how that goes the other way (trig)
obtuse pebbleBOT
how is the sum to product rule derived? I get the product to sum rule, but I can...
cunning burrow
how is the sum to product rule derived? I get the product to sum rule, but I can...
Can you write the theorem down
thorn summit
Can you write the theorem down
too lazy to write so here it is
cunning burrow
too lazy to write so here it is
So why do you want to prove the other way?
cunning burrow
"" but I can't understand how that goes the other way (trig)""??
thorn summit
the only proofs i've seen use the product to sum rule to derive each one
no like
i get this set of identities
but i don't know how you go from this ^ to the other set
cunning burrow
I see
cunning burrow
too lazy to write so here it is
Why not divide by 2 and then create two variable u and v
thorn summit
Why not divide by 2 and then create two variable u and v
2sin(a)cos(b) = sin(a+b) + sin(a-b)
now what
cunning burrow
now let u =a+b and v = a-b
thorn summit
2sin(a)cos(b) = sin(u) + sin(v)
cunning burrow
For the LHS also make the substitution
thorn summit
WAIT
I JUST FINALLY REALIZED
OHH
2sin((u+v)/2)*sin(u-v/2) = sin(u) + sin(v)
because (u-v)/2 is equal to b, and (u+v)/2 = a
is that right?
whole dock
thorn summit
yeah i meant cosine mb lmao
thorn summit
thank you that actually makes sense now
the only reason i was confused was why you add the two angles
ill close the thread now
ty for the help
saved me from bashing my head into a wall
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
lofty mason
can someone explain this
obtuse pebbleBOT
can someone explain this
lofty mason
<@&286206848099549185>
obtuse pebbleBOT
@lofty mason Has your question been resolved?
lost otter
obtuse pebbleBOT
@lofty mason Has your question been resolved?
lofty mason
<@&286206848099549185>
fluid holly
warm shaleBOT
lofty mason
thank you
obtuse pebbleBOT
Available help channel!
fallow widget
$\int \frac{2x^2-4}{x^3+2x^2} dx$
obtuse pebbleBOT
$\int \frac{2x^2-4}{x^3+2x^2} dx$
warm shaleBOT
Merineth
fallow widget
I'm having a really hard time figuring out what to do with integrals when i have problems such as these
I know that my options are partial fraction, integration by parts and long division
However is there a way to more easily recognize which one to do?
timid silo
have you tried Numerator = Ad/dx(denominator) + B(denominator)
warm shaleBOT
deltaG
past sand
First thing to do would be to factor things as much as possible
Then, since you have a higher degree in the denominator, it's not going to be long division, it's likely going to be partial fractions
robust raven
**Merineth**
fallow widget
$\int 2\frac{x^2}{x^3+2x^2}dx -4\int \frac{1}{x^3+2x^2}dx$
Yeah but i'm not asking for the answer
I'm asking for a method for how to figure out myself on which method to choose
As i can't see which one that will lead me to an answer.
@past sand Also, is this what you meant? And now factor denominators?
warm shaleBOT
Merineth
robust raven
in your example there is a proper rational function, i.e. the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator, this is how such expressions integrate, sometimes the derivative of the denominator is included in the numerator, but you won't get it effectively here
past sand
<@525582731569266698> Also, is this what you meant? And now factor denominators?
No you don't have to split
fallow widget
But how do you know i don't have to split?
I would've naturally split them to make it easier
past sand
Factoring is just to make sure you don't have obvious simplifications and also to set up partial fractions
fallow widget
Since i couldn't do u-sub on it to cancel out the numerator
past sand
Well you can split, it's just not necessary
I guess it's easy here since the 2x^2 in left term cancels out and you're left to do partial fractions on the right term
What I meant by factoring was just getting to this: 2(x^2-2)/(x^2(x+2))
Then you put the constant 2 outside, split the x^2-2 if you want, and do partial fractions
fallow widget
I can't see it
$\int \frac{2x^4+3x^3-4}{x^3+2x^2} dx = 2x-1 + \int \frac{2x^2-4}{x^3+2x^2}dx$
This was my original that i got
i applied long division on it as i noticed my numerator had a higher exponent than my denominator
But i can't for the love of me figure out what to do now with the remaining integral
warm shaleBOT
Merineth
past sand
The 2x-1 part is good but still needs to be integrated
fallow widget
$\int \frac{2(x^2-4}{x^2(x+2)}dx \implies \frac{2(x^2-4}{x^2(x+2)} = \frac{A}{x^2} + \frac {B}{x+2}$
warm shaleBOT
Merineth
fallow widget
Is this how its done?
past sand
Almost
There's a third term, but maybe try to find A and B just from this
Also careful with that parenthesis
fallow widget
Well i got B = 0 and A = -4
$2(x^2-4) = A(x+2) + Bx^2$
If x = -2 \
$0 = 4B \implies B = 0$
warm shaleBOT
Merineth
fallow widget
It's right tho?
robust raven
**Merineth**
thatis not correct decomposition
fallow widget
$2(x^2-4) = A(x+2) + Bx^2$
warm shaleBOT
Merineth
past sand
It's not 2(x^2-4)
.
fallow widget
$2((-2)^2 - 4) = A(-2+2) + B(-2)^2$
warm shaleBOT
Merineth
robust raven
**Merineth**
do you really not understand that your decomposition is incorrect?
you should have sum of three fractions, not two !
and that is not calculus but algebra
Nel was saying it to you too
past sand
you should have sum of three fractions, not two !
That's the next step, no need to rush
robust raven
That's the next step, no need to rush
i see ok
fallow widget
$2(x^2-2) = A(x+2) + Bx^2$
warm shaleBOT
Merineth
fallow widget
I got:
B = 1
A = -2
past sand
**Merineth**
Expand this and show us how you get A and B
warm shaleBOT
Merineth
past sand
that's not how it works
fallow widget
Yes it is?
past sand
You can't just take any value of x
fallow widget
Ofc i can?
past sand
... no
fallow widget
That's how i've done it for the past 4 weeks
past sand
If you take a specific value of x you're only proving your decomposition works for that value of x
It doesn't prove at all that 2(x^2-2) = -2(x+2) + 1x^2
fallow widget
"or you can plug in x values and determine A and B"
That is quite literally what i'm doing
past sand
Oof that's such bad advice
unreal musk
As in use the values that you found for A and B 
fallow widget
Anyhow even if i solve this partial fraction integral
it doesn't really answer my original question
unreal musk
Oof that's such bad advice
(It isn’t when it comes to finding the constants, if it holds for any choice of x you can choose whatever values you want, no?)
fallow widget
I am unable to see the pattern of which method to choose
so even if a similar question arrived on the exxam on monday
I wouldn't be able to sovle it because i can't find the pattern on what method to choose
the only method i'm 100% that i knew i could do was Long division since the exponent difference in numerator/denominator
past sand
(It isn’t when it comes to *finding* the constants, if it holds for *any* choice...
Not sure what you mean, even the example in the video has a single solution that's really easy to find as a system of two equations
versed stratus
notice this is $2(x^2-2)/x(x^2+2)$ add and subtract four to the numerator and then use partial fractions
warm shaleBOT
Why am. I here
versed stratus
*4 not 2
versed stratus
oh, didn't notce, sorry
past sand
the only method i'm 100% that i knew i could do was Long division since the expo...
Numerator degree >= denominator degree --> long division
Otherwise, if it's still complicated, try partial fractions
If it's simple, you may already know an antiderivative or a classic substitution
robust raven
**Merineth**
unreal musk
**Merineth**
As in something like this holds for all values of x, you can choose any value of x and it must give you a true statement, which forces you to know what A and B are
past sand
I wouldn't be able to sovle it because i can't find the pattern on what method t...
While I agree, knowing you should do partial fractions won't help if you can't do it properly
fallow widget
$\int \frac{2(x^2-4)}{x^2(x+2)}dx \implies \frac{2(x^2-4)}{(x)(x)(x+2)} = \frac{A}{x} + \frac {B}{x+2} + \frac{C}{x}$
warm shaleBOT
Merineth
unreal musk
Choosing values of x will get you the solutions just as e.g. equating coefficients would
past sand
Alright do you want a solution for this one?
past sand
(woah Discord getting laggy)
versed stratus
**Merineth**
Uh, there's a simplification you can perform here, instead of going straight for partial fractions(Hope your key mentioned that)
fallow widget
my key?
unreal musk
Uh, there's a simplification you can perform here, instead of going straight for...
(afaik the top shouldn’t even be that 
)
past sand
I already have the solution
You mean you have the teacher's solution?
fallow widget
Yes.
but i couldn't solve it on my own
Because i can't
- See what method to choose
- How the partial fraction becomes so fucked
past sand
Alright do you want a solution for this one?
Well yeah it's not your own, that's why I'm asking that
past sand
Numerator degree >= denominator degree --> long division
Otherwise, if it's stil...
For the first point
past sand
Yeah that's what I wanted to get to but I didn't know you learned to solve these by "choosing a value for x", which to me is kind of insane
fallow widget
System equations?
past sand
Yes
Let's try this again shall we?
$\frac{2x^2-4}{x^2(x+2)} = \frac{A}{x^2} + \frac{B}{x+2}$
warm shaleBOT
Nel
past sand
$2x^2-4 = A(x+2) + B(x^2)$
warm shaleBOT
Nel
warm shaleBOT
Nel
unreal musk
Yeah that's what I wanted to get to but I didn't know you learned to solve these...
(Tl;dr It does work though)
past sand
The thing to do now is to match each of the different degrees of x together
fallow widget
So b is 2?
past sand
Do you see how B matches with 2, and 2A matches with -4?
Yeah, however we have a problem
If 2A = -4 that means A = -2 and the Ax doesn't match with anything
fallow widget
Do you see how B matches with 2, and 2A matches with -4?
2A matches with -2? -4*2 would be -8
past sand
It's -4 on the left
past sand
Yes
fallow widget
Ok
past sand
So you end up with -2x on the right side and nothing like it on the left side
fallow widget
So Ax has to be 0 for the equation to match
past sand
Yeah but that would mean x=0 which is not something we want, this should work for any value of x
This just means the decomposition is incorrect
fallow widget
I see
past sand
In short, to fix that decomposition, you need a third term C/x
That will give you some more x to solve with, and match the -2x
So: $\frac{2x^2-4}{x^2(x+2)} = \frac{A}{x^2} + \frac{B}{x+2} + \frac{C}{x}$
warm shaleBOT
Nel
past sand
$2x^2-4 = A(x+2) + B(x^2) + Cx(x+2)$
fallow widget
I see what you are trying to say but i'm having a really hard time accepting it
warm shaleBOT
Nel
fallow widget
Yes but
I wouldn't be able to solve a similar question
I want to understand how we get C/x
past sand
Well, here's the rule: once you have factored the denominator as much as possible, you get a product of polynomials of various degrees
past sand
**Nel**
Here it's two polynomials: x^2 and x+2
The first has degree 2, the second has degree 1
For each of these, you need as many partial fractions as the degree number
fallow widget
Okay that i can remember for sure
past sand
Kind of hard to explain
fallow widget
I totally get what you mean
past sand
past sand
Yes
fallow widget
A/x^3 + B/x^2 + C/x
fallow widget
And to confirm that i did it right we check for LHS = RHS
past sand
Yes well you solve for each of your constants A, B, ...
fallow widget
Yes
past sand
If there's something leftover then it doesn't work
fallow widget
Okay that makes sense, i'm pretty sure i'll be able to solve them now.
No wonder i was struggling so much with those, the book only taught / gave examples that had the exponent ^ 1. Not a single example of x^2, or higher than x^1
past sand
Btw instead of 2x^2-4 you can work with x^2-2 and keep the factor of 2 outside of the integral
fallow widget
Just to double check
past sand
Might be easier, might not be, but sometimes making it as simple as possible can show you simplifications and stuff
fallow widget
$(x^2 - 2) = A(x+2) +Bx^2 + Cx(x+2)$
warm shaleBOT
Merineth
unreal musk
Most likely different labelling than you picked
past sand
Also, sometimes you get irreducible quadratics in the denominator
Like f(x)/[x^2(x^2+5)(x-4)]
The (x-4) is simple, the x^2 you know what to do, and then you have (x^2+5) that you can't factor
That just means you decomposition will have a term D/(x^2+5)
unreal musk
That just means you decomposition will have a term D/(x^2+5)
Dx + E you mean right?
past sand
Oh yeah
fallow widget
$-2\int\frac{1}{x^2}dx$
warm shaleBOT
Merineth
unreal musk
This one is really easy 
past sand
Dx + E you mean right?
(I've said it before but I'm bad at maths)
unreal musk
my feelings
awwww 
past sand
This is just the power rule
unreal musk
Yea that 
fallow widget
the power rule?
past sand
(x^n)' = nx^{n-1}
x^n is always the power rule, except for n=-1 (so 1/x) where the antiderivative is ln(x)
fallow widget
unreal musk
You have your formula thingy don’t you 
past sand
Ye
warm shaleBOT
Merineth
fallow widget
,w integrate (2x^4+3x^3-4)/(x^3+2x^2)
unreal musk
2x^4
fallow widget
hihi
warm shaleBOT
unreal musk
And you forgot to integrate the first two 
past sand
Yeah you forgot to integrate 2x-1
fallow widget
:)
past sand
You did long division inside the integral, so the result stays inside
fallow widget
Yeah completely forgot about it
did the long division like 1.5 hours ago lmfao
oufh
I haven't practiced second order ODE
$y''(x) + 4y(x) = -36sin(4x) -60cos(4x)$
warm shaleBOT
Merineth
fallow widget
I know i want the yh(x) first to find yp(x)
But first of all i need to write it in terms of y'' +ay'+by =f(x)
past sand
And that's my cue to leave 
fallow widget
i'm missing a ay' ?
unreal musk
But first of all i need to write it in terms of y'' +ay'+by =f(x)
(which you kinda do already can you identify the stuff you need to? 
)
unreal musk
i'm missing a ay' ?
"missing" so what could a be?
unreal musk
Well, a = 0
(if y' = 0 then y'' = 0 and, well, life would be a lot easier to deal with 
)
warm shaleBOT
Merineth
fallow widget
oh i think i remember now
it's coming back to me!
I write it in terms of
r^2 +0r +b
And then factorise it?
r^2 + 4
hm
unreal musk
And then factorise it?
(it may not be easily factorable)
fallow widget
clearly D:
unreal musk
This one is "somewhat" easy to factorise though, if you know where to look
unreal musk
(anyways, to also kinda make it maybe clearer to see, what's most important is that you want the roots to the polynomial $r^2 + ar + b = 0$, and those are your $r_1$ and $r_2$ they refer to)
warm shaleBOT
@unreal musk
fallow widget
x+y = 0
x*y = 4
I tried this but if i make y = -x i get a imaginary number
since x^2 = -4
unreal musk
Which you should! 
unreal musk
Almost, not quite 
fallow widget
really?
unreal musk
since x^2 = -4
what are the solutions to this?
unreal musk
imaginary number
It is, still no idea what specific imaginary number(s) they become?
fallow widget
I don't have a clue
unreal musk
r^2 + 2r +2r + 4 -4r = r^2 + 4
And well I see you edited 
but that's not really "factored" because that -4r is separate?
unreal musk
Do you know the solutions to x^2 = -1?
fallow widget
It is, still no idea what specific imaginary number(s) they become?
x = 2i ?
No
That was purely a guess
I honestly had no idea
unreal musk
x = 2i ?
That's right and you also need x = -2i too alongside, solutions come in conjugate pairs!
unreal musk
You can also notice that r^2 + 4 is a difference of two squares, but replacing -i^2 = 1 to get r^2 - 4i^2 = r^2 - (2i)^2
warm shaleBOT
Merineth
fallow widget
where do you get
uh
i have no idea what you just did
anyhow
they are not equal
r_1 and r_2
So i use the top one
unreal musk
Welllllll, you coulddddd, buttttt
fallow widget
It says in the formula that
unreal musk
fallow widget
if the roots are
aaah fuck
to slow
If the roots are complex conjugate to each other i can use that formula (which they are)
unreal musk
Easier to use that form, though technically the top one does work
fallow widget
What happens with y_p(x) ?
unreal musk
Well now, we have are homogenous solutions, with alpha = 0, beta = 2
fallow widget
$y_h (x) = e^{\alpha x}[Acos(\beta x) + Bsin(\beta x)]$
warm shaleBOT
Merineth
unreal musk
Did you happen to have anything that explained the particular solution and how you choose them?
fallow widget
yp(x) ̈is an arbitrary particle ̈solution to the equation P (D)y(x) = f (x) obtained via suitable approach.
unreal musk
hmmm, very well explained 
fallow widget
**Merineth**
what is alpha, beta, A and B ?
unreal musk
what is alpha, beta, A and B ?
They're the complex solutions, here you had your solutions as 0 ± 2i
unreal musk
that the solutions were 0 ± 2i, or that alpha = 0 and beta = 2?
unreal musk
r_1 = +2i
r_2 = -2i
so r1 is 0 + 2i and r2 is 0 - 2i, complex conjugate pairs...
past sand
fallow widget
$\alpha = 0$ ?
warm shaleBOT
Merineth
past sand
No
fallow widget
why zero then ?
unreal musk
Not necessarily, or at least a very bad way to think about it
past sand
Because r_1 = 2i = 0 + 2i
unreal musk
Think about alpha as the real part of one of your solutions, and beta as the imaginary part
past sand
Wdym factoring?
fallow widget
When factoring r^2 + 4
we got pm 2i
is alpha always 0
or is it possible that we get Re(z) = something other than zero
past sand
Of course alpha can be non-zero
I guess when the coefficient of y' is zero then alpha is zero too, but that's not the point
unreal musk
I guess when the coefficient of y' is zero then alpha is zero too, but that's no...
[and the coefficient of y strictly positive - to get a complex non-real solution]
past sand
There is no i in there
unreal musk
2x
fallow widget
ah
unreal musk
And you don't need the ± there either
warm shaleBOT
Merineth
past sand
or is it possible that we get Re(z) = something other than zero
As an example, y''+2y'+5y would give you r^2+2r+5 = (x+1+2i)(x+1-2i)
Here you'd have alpha = 1, beta = 2 or -2
past sand
Never mind
fallow widget
that seems very hard to factor
now sure i'd be able to do it
x+y = 2
x*y = 5
y = 2-x
x*(2-x) = 5
unreal musk
(remember that you can just instead solve r^2 + 2r + 5 = 0 and then choose r1 and r2 as the solutions to that)
warm shaleBOT
Merineth
fallow widget
$-1 \pm \sqrt{-4}$
past sand
Why -1?
warm shaleBOT
Merineth
fallow widget
-1 pm 2i
past sand
Oh wait does alpha = -1 work?
unreal musk
,w solve r^2 + 2r + 5 = 0
warm shaleBOT
unreal musk
So it does 
unreal musk
As an example, y''+2y'+5y would give you r^2+2r+5 = (x+1+2i)(x+1-2i)
(cause here you're doing (x - r1)(x - r2) which makes the -1 become +1)
past sand
y''(x) + 0y'(x) + 4y(x) = -36sin(4x) -60cos(4x)
It's going to be some Csin(4x) + Dcos(4x)
fallow widget
past sand
Since the derivatives of sin and cos just keep being sin and cos (with some negatives), both y'' and y being of the same form as the RHS works
fallow widget
Oh so we guess what the particular solution is going to be based on the RHS of the equation?
past sand
Same thing with e^{kx}, like y'' + 4y = 3e^{4x} or something
fallow widget
Do you have any more examples i could try?
past sand
Same thing with e^{kx}, like y'' + 4y = 3e^{4x} or something
Well you could try this
unreal musk
tl;dr yes (one thing to check/convince yourself of, is that the RHS is not a solution to the homogenous version, and if it is, "multiply by x until and only until it isn't" is the general approach)
fallow widget
unreal musk
E.g. for the polynomial term, you're gonna guess that it's some general quadratic polynomial (if it doesn't form a solution to the homogenous version)
unreal musk
What's the guess you're making for the particular solution?
fallow widget
$y_p (x) = Ae^{2x} -Be^{2x} +Ce^{2x}$
warm shaleBOT
Merineth
unreal musk
Erm, I mean, kinda...? You just need one coefficient
You only need to think about Ae^{2x} - buuttttttt...
Remember what I said about checking whether you're a solution of the homogenous equation first?
warm shaleBOT
Merineth
unreal musk
Yea it's supposed to be the other way around, remember it's (r - r1)(r - r2)
warm shaleBOT
Merineth
unreal musk
Well, not quite
unreal musk
**Merineth**
More that the homogenous solution y'' - 3y' + 2y = 0 is this, but the reason we cared about that was...
Seeing that C1 and C2 are arbitary, we could choose C1 = 1, C2 = 0
unreal musk
You only need to think about Ae^{2x} - buuttttttt...
...wait, we can? Shit
fallow widget
$y_p (x) = Ae^{2x}$
warm shaleBOT
Merineth
unreal musk
It's actually $y_p(x) = Ax e^{2x}$
warm shaleBOT
@unreal musk
fallow widget
huh
unreal musk
Remember the whole thing I said about "multiply by x until and only until it isn't"?
fallow widget
unreal musk
Check if they mention anything about being a solution of the homogenous solution
unreal musk
chatGPT being "useful" [sic] again 
fallow widget
$y(x) = e^{0x}[Acos(2 x) + Bsin(2 x)] + Ce^{2x}$
warm shaleBOT
Merineth
unreal musk
Why are there no sources that explain this well I can find 
fallow widget
woops
i mixed my questions
$y(x) = e^{0x}[Acos(2 x) + Bsin(2 x)] + Csin(4x) + Dcos(4x)$
warm shaleBOT
Merineth
fallow widget
there we go?
unreal musk
Well that's better, but first
We want to find the values of C and D
They can be determined
Let $y_p(x) = C\sin(4x) + D\cos(4x)$, and that it's a solution to the original differential equation, so that $y = y_p$ satisfies $y'' + 4y = -36\sin(4x) - 60\cos(4x)$
warm shaleBOT
@unreal musk
fallow widget
By putting in the particular solution inside the y_h(x) ?
Idk this is getting very complicated and very messy
I'm not sure where i am atm
unreal musk
Nooo, you put the particular solution into the original differential equation then e.g. equate coefficients
fallow widget
Isn't there a step by step that i could remember?
- find yh
- find yp
- put yp into og eq to find constants
.. then?
unreal musk
I've been trying to find a good explanation of one (but haven't seen one yet 
)
unreal musk
* find yh
* find yp
* put yp into og eq to find constants
But yep, that's kind of it, then once you find the coefficients of the particular one, you can add them together at that point
So, for the finding the coefficients, I'm sure you're happy with finding the second derivative of $C\sin(4x) + D\cos(4x)$, right?
warm shaleBOT
@unreal musk
fallow widget
unreal musk
And this becomes?
unreal musk
naughty using = there, but I get what you mean, and that's fine
unreal musk
Yep, that's it
So you get
[
-16A\sin(4x)-16B\cos(4x) + 4A\sin(4x) + 4B\cos(4x) = -36\sin(4x) - 60\cos(4x)
]
warm shaleBOT
@unreal musk
unreal musk
You can hopefully simlplify that down
unreal musk
I'll take your word for that now
warm shaleBOT
Merineth
fallow widget
I just compare coefficients
unreal musk
So, you'd have $y_p(x) = 3\sin(4x) + 5\cos(4x)$, then adding $y_h(x) = C_1 \cos(2x) + C_2 \sin(2x)$ to that gets you your complete solution
warm shaleBOT
@unreal musk
unreal musk
And now, our moment of truth, we should find that $y(x) = y_h(x) + y_p(x) = C_1 \cos(2x) + C_2 \sin(2x) + 3\sin(4x) + 5\cos(4x)$
warm shaleBOT
@unreal musk
fallow widget
**Merineth**
this
unreal musk
,w y'' + 4y = -36sin(4x) -60cos(4x)
warm shaleBOT
unreal musk
didn't we use
e^{0x} is 1
fallow widget
$y(x) = e^{0x}[3cos(2 x) + 5sin(2 x)] + 3sin(4x) + 5cos(4x)$
warm shaleBOT
Merineth
fallow widget
oh
unreal musk
The constants of cos(2x) and sin(2x) aren't determined, the ones of sin(4x) and cos(4x) are
warm shaleBOT
Merineth
fallow widget
Okay so
- Find yh
- Find yp
- plug yp into og equation to find constants
- add yh + yp
win
unreal musk
Yep, that's the idea
(I'll have to still find something that explains the particular solution, but remember, tl;dr make sure your original f(x) is not a solution to the homogenous equation!)
If you go down the page, they explain the whole "make sure you're not a homogenous solution" thing I was saying
fallow widget
♥️
I got a quick question chart
before you flee
I'm supposed to draw this
i already figured out the y'
but asymptote
Still struggling a little
unreal musk
fallow widget
unreal musk
The vertical one(s) should be hopefully easy to find...
fallow widget
I see that
one asymptote
yeah
is -1
but to check for oblique ones
i do
lim x -> infinity (f(x) -kx-m) where k = f(x)/x
i tried to find k but
i got it to be infinity?
$y = (\frac{x}{x+1})^2 = \frac{x^2}{(x+1)^2} = \frac{x^2}{x^2+2x+2}$
warm shaleBOT
Merineth
fallow widget
Right?
unreal musk
+1 but sure 
fallow widget
oh yea lol mb
anyway
i didivde this by x now
which is equivalent to multiplying the denominator with x
right?
$\frac{x^2}{x^3+2x^2+x}$
warm shaleBOT
Merineth
fallow widget
If x = 0 now i would get infinite large?
unreal musk
Well if x = 0 sure, but remember you want x to go to infinity
Also notice the numerator and denomiantor have a common factor...
fallow widget
OH
factor and cancel one
x
damn
$\frac{x}{x^2+2x+1}$
Now we have to check for m?
unreal musk
Not quite, almost 
warm shaleBOT
Merineth
unreal musk
fallow widget
Wouldn't this result in it becoming infinite small?
-infinity
since x^2 is faster than x
unreal musk
Well the whole thing goes to zero
fallow widget
Oki
But wasn't the condition for an oblique asymptote
that k and m has to be
actual numbers
and not infinity
unreal musk
**Merineth**
You do get actual numbers, this goes to zero when x goes to infinity...
fallow widget
Meaning we can conclude that there are no oblique asymptotes
unreal musk
since x^2 is faster than x
...for this reason here
fallow widget
But we can still draw the conclusion about there being no oblique asymptotes?
unreal musk
Well if there were one, it would be horizontal, sure
fallow widget
uh
unreal musk
Now it's worth checking whether there's a horizontal one
fallow widget
Now it's worth checking whether there's a horizontal one
is that what m is for?
unreal musk
is that what m is for?
Yep, you (should be happy!) that k = 0, so now it's finding m
unreal musk
(via the first part of this here!)
fallow widget
m = lim x->infinity f(x)-kx
So since k = 0
kx is 0
meaning we only have f(x) left over
$m = \lim_{x \to \infty} (\frac{x}{x+1})^2$
unreal musk
\lim_{x \to \infty}
idk how to lim
warm shaleBOT
Merineth
warm shaleBOT
Merineth
warm shaleBOT
Merineth
unreal musk
no and no
fallow widget
hmm
unreal musk
First off, you want x really big
fallow widget
well the larger x becomes
numerator will never be larger than denominator
since we have the constant +1
in the denominator
denominator will always be +1
unreal musk
Hmm, while a factually true statement, notice the numerator and denominator have the same degree...
unreal musk
Basically, kinda
unreal musk
**Merineth**
But like if there's an x^2 at the top, and an x^2 at the bottom...
fallow widget
They cancel each other out?
leaving a 1 at the numerator?
1/1
= 1
so we have a horizontal asymptot at y = 1
?
unreal musk
Or alternatively $\frac {x^2}{x^2 + 2x + 1} = \frac1{1 + \frac2x + \frac1{x^2}}$
warm shaleBOT
@unreal musk
obtuse pebbleBOT
unreal musk
Need a food break now...
Enjoy the food 
obtuse pebbleBOT
Available help channel!
mint nebula
What's the product of all the real solutions of the ecuation
obtuse pebbleBOT
What's the product of all the real solutions of the ecuation
mint nebula
obtuse pebbleBOT
brazen gorge
what are [] and {} ?
velvet wyvern
Multiply the equation by (x^2+3)
zealous bobcat
what are [] and {} ?
integer part and fracional part
tepid falcon
Can someone help me solving this?
obtuse pebbleBOT
Can someone help me solving this?
wooden cipher
!original
obtuse pebbleBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
tepid falcon
The full context is here, let me translate it
It basically says that we have a equation for the shear strength
The equation (1
Then we should replace equation 2) that reffers to Newton's viscosity law
tepid falcon
Can someone help me solving this?
And then we have this equation
tepid falcon
The full context is here, let me translate it
The one that is between equation 1) and 2) in this photo
Should i provide more context?
tardy epoch
Can someone help me solving this?
Are you just solving for v_z(r)?
tepid falcon
The full context is here, let me translate it
Sure, I forgot to mention that the result should be equation 3)
tepid falcon
Are you just solving for v_z(r)?
Yes
tardy epoch
Can someone help me solving this?
You integrate this equation both sides by r first. Then divide by x-mu * r to isolate just v_z'(r) and then integrate again
tepid falcon
You integrate this equation both sides by r first. Then divide by x-mu * r to is...
Which metod are you using
warm shaleBOT
riemann
tepid falcon
You supposed to solve it by a differential equations metod, am I right?
tardy epoch
You supposed to solve it by a differential equations metod, am I right?
Integration is a valid method for solving differential equations
tepid falcon
Integration is a valid method for solving differential equations
So you mean i should use variable separation?
tepid falcon
You should just do what I said
Could you tell me how would be the equation before integrating it?
tardy epoch
Can someone help me solving this?
Just integrate the right side with respect to r
tepid falcon
Just integrate the right side with respect to r
You mean this?
tardy epoch
Yes
obtuse pebbleBOT
@tepid falcon Has your question been resolved?
tardy epoch
I think i'm wrong
Well you have to figure out what the constant is
tardy epoch
The full context is here, let me translate it
There's no C in (3
obtuse pebbleBOT
@tepid falcon Has your question been resolved?