Lokisan! Please help me with more differential equations! :)

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Lokisan! Please help me with more differential equations! :)

Mathematics
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22xdy+11ydx+3ydy=0
Loki! haha I'm back with more questions. :)
Hehe, I see...

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Other answers:

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?
Hang on...I misread it...I have to work something out. Sorry, I'll just finish with this other person and I won't be distracted.
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
I keep being called away! I'll just let you know what I'm thinking and you can play with it as well. Your equation, when rearranged, looks like this\[y'=-\frac{11y}{22x+y}=-\frac{11(\frac{y}{x})}{22+(\frac{y}{x})}\]after dividing the numerator and denominator by x.
This is in the form for a type of differential equation that can be solved when we know that y' is some function of y/x; that is,\[y'=F(\frac{y}{x})\]
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?
You have a second order differential equation which is linear, inhomogeneous (because the RHS is a function of x only) with constant coefficients. You need to solve inhom. equations by 1) solving the equivalent homogeneous equation first 2) finding a particular solution
+
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
Thanks so much... I really feel low just asking for answers but I plan on taking the class again. If I fail I have to pay the whole tuition and I've been studying so hard and just not getting it lately :(
It's because you're having to do it on your own - you're not being shown the ropes. Don't punish yourself.
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.
\[4xydx=(x^2+4)dy \rightarrow \frac{4xy}{x^2+4}=\frac{dy}{dx} \rightarrow \frac{4x}{x^2+4}dx=\frac{dy}{y}\]
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...
You don't have to tell it to solve for y; it guesses. If it doesn't, then it's a form it doesn't recognize.
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?
Maybe she's getting you ready to think in terms of differentials or something. Really, it doesn't matter. Just stick to what she does.
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.
Good luck then. If you need help, look up an old post and send something. I'll get an e-mail about it.
Ok I will man. thanks. Hopefully if you're around I can ask for your help like this again I would star you 1000 times if I could
No worries. have a go at that first one. The method of substitution v=y/x should work. You end up with a differential equation in v which is separable...and you're comfortable with that ;)
I'm going to go offline for a while. I have some stuff to do. Good luck!

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