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22xdy+11ydx+3ydy=0

Loki! haha I'm back with more questions. :)

Hehe, I see...

Are you studying exact differential equations?

I don't think so. I know what the answer is and its not a number. Its the formula of the constant

Can you wait a little bit - I'm just finishing another post.

Sure

22xdy+11ydx+ydy=0 is the actual problem, Sorry

ok

i need paper

haha alright

oh, i don't have her question.

Scotty, this looks worse than it is (sorry, I have someone else I'm helping here too).

I'm hoping so. I think it fits into the trick of d(xy)=dy+ydx am I correct?

Ok so the problem is 22xdy+11ydx+3ydy=0

I wish that 22 wasn't there. You sure it's not 11?

Positive :/

I know what the answer if, if that helps.

yeah punch it in

Its \[11xy^2+y^3=C\]

Sorry, I keep being called away.

It's alright..

I'm thinking a substitution method. Just one sec...again called away...sorry. Won't be forever.

I think I may have your method...just let me check properly.

Ok cool

We make the identification,\[v(x)=\frac{y}{x}\]so that\[y=xv\]

and we take it from there. Is this familiar?

Ok let me try to pmess with it

Ok instead of that one... can you help me solve y'-9y=8e^x

I'm doing this one for a timed quiz

I figured out your other problem - I made an arithmetic error, but the method is correct.

Yeah I'll do the second.

Ok I think I saw where you were going with it and I gutted my way through it. Thanks

Okay, integrating factor one - do you know?

Not sure what that means..

\[y'-9y=8e^x \rightarrow \frac{d (\mu y)}{dx}=\mu 8e^x\]

\[\mu = e^{\int\limits_{}{}-9dx}=e^{-9x}\]

so

im sorry that should be
y"

\[e^{-9x}y=e^{-9x}.8e^x+c \rightarrow y=ce^{9x}-e^x\]

y''-9y=8e^x ?

Yes, sorry for the confusion. I miss typed.

right
do you just want the answer for the quiz and then the method

Well. I want to learn but first and foremost I really need to pass this quiz

I am writing it all down to try to understand though.

\[y=c_1e^{3x}+c_2e^{-3x}-e^{x}\]

I'll explain...

Are you sure the + isn't a - between the first 2 terms?

In the end, it doesn't matter because it's absorbed into the constant.

The only option I have with a + sign in it is...
-2e^3x+e^-3x-e^x

Ok so e^3x=e^-3x-e^x is the same thing?

Oh, do you have initial conditions?

Sorry when x=0

Your equation has it's constants solve for

Oh...you need to give me the whole question ;)

Wait.... y=0 and y'=8 and x=0... I'm sorry I've been up all day doing this

It's second order, so there should be 2 conds.

ok

c_1 = 2 and c_2 = -1

\[y=2e^3x-e^{-3x}-e^x\]

Yep that works!

Good

trying not to. Could you help me with 1 or 2 more? Then I'll leave you alone.

yes

How long do you have to do the quiz?

i panic in tests - you shouldn't have told me, lol

4xydx=(x^2+4)dy
and that is the whole problem and not leaving anything out :)

It's separable.

You can integrate from there.

Ok let me try

\[y=c(x^2+4)^2\]

yep thats what I got

Aesome

or, awesome

Ok I think this should be the last one... I think I got the others actually.

ok

dy+2ydx=e^(-4x)dx

It's an integrating factor one again.

Yeah those are really difficult for me. Seperatable ones are the ones I'm more comfortable with...

\[y=-\frac{1}{2}e^{-4x}+ce^{-2x}\]

I'm just skipping to the solution because of the test...

The closest answer I have that looks like that is e^(4x)+Ce^(2x)

hmmm...that's the solution to the equation you gave me...

let me write it in the formula so it looks better maybe it looks funny on text

\[dy+2ydx=e^{-4x}dx \rightarrow y'+2y=e^{-4x}\]?

\[dy-2ydx=6e^(5x) dx\]

that dan
5x is super script

so you gave me the wrong question before?

Im a failure sorry I'm rushing

\[y(x) = c_1 e^{2 x}+2 e^{5 x}\]

is that one available?

yes just not written that way but yeah

I had actually guesed that one after trying to work it myself. Hurray for small miracles.

so you've passed?

Well... not 100% sure. I think I got the rest of them right.

Do you still want to look at the first one?

For dy+2ydx=e^(-4x)dx I got.... y=-1/2e^(-4x)+ce^(-2x)

that's what I got..?

and thats the only other one I kind of guessed on. I do want to look at those though yes.

No I did that one myself

good

I think its right, not 100% sure though

It's right

if

it's the one I did before...and it is...it's right.

You're fine at this!

Oh ok, sorry I'm scattered. ok I'm submitting it!

What about your first question?

How much time do you have?

5 mins left

the first one was a practice but I do want to know how to do it

ok

How about....
is y'=2y(x-3) seperable

I said yes.

yes, separable

ok and 2xdx+ysinxdy=sinxdx

I got y^2=x-2sinx+C

that's separable too - is that how you solved it?

Yes

that one would have taken a bit of effort

Ok is y'=y/x y=cx

yes

ok well it looks all good then :)

Thank you so much for that man.....

that's okay - i hope you pass

have you heard of a site called wolframalpha?
www.wolframalpha.com

Ok so that original problem..

You can type in things and it will calculate, like your diff. eqs

but it won't show you what to do, that's the only thing

I have, I don't know how to type these kinds of problems in though.... could you show me how?

I'm just thinking, if you have more of these tests, you can have a look at that as a resource

I did use it before, I have 2 more tests actually.

y'-2y=e^(-4x)
is what you can type in for one of your d.e.'s above. It will give you the answer.

Use primes on the y's...

Hmm.... but even for stuff like... 22xdy+11ydx+3ydy=0?

Yeah, divide everything on your paper by dx so you can have (dy/dx) which you can equate as y'

lol... thats a genius way to do it.

\[22x \frac{dy}{dx}+y+3y \frac{dy}{dx}=0 \rightarrow 22xy'+y+3yy'=0\]

I don't know why they keep giving you the equation in that form...it's weird.

In the form I gave you? Thats the way my teacher gives me them....

Yeah, I know...weird.

Almost all my questions are like that is that not how they are normally done?

lol, no

Hmm why would she try to teach like that then?

did you pass this quiz?

Ok well I really appreciate it. You might be the only reason I passed that quiz... I'm sure of it.

I won't know for a day or two, but I'm sure I did.

Cool, hope so...don't want to be spending more money on tuition...

god I know. I have...2 more quizes and a final. Then I'll be done with this class.

I'm going to go offline for a while. I have some stuff to do. Good luck!