anonymous
  • anonymous
Lokisan! Please help me with more differential equations! :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
22xdy+11ydx+3ydy=0
anonymous
  • anonymous
Loki! haha I'm back with more questions. :)
anonymous
  • anonymous
Hehe, I see...

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anonymous
  • anonymous
Are you studying exact differential equations?
anonymous
  • anonymous
I don't think so. I know what the answer is and its not a number. Its the formula of the constant
anonymous
  • anonymous
Can you wait a little bit - I'm just finishing another post.
anonymous
  • anonymous
Sure
anonymous
  • anonymous
22xdy+11ydx+ydy=0 is the actual problem, Sorry
anonymous
  • anonymous
ok
anonymous
  • anonymous
i need paper
anonymous
  • anonymous
haha alright
anonymous
  • anonymous
oh, i don't have her question.
anonymous
  • anonymous
Scotty, this looks worse than it is (sorry, I have someone else I'm helping here too).
anonymous
  • anonymous
I'm hoping so. I think it fits into the trick of d(xy)=dy+ydx am I correct?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Ok so the problem is 22xdy+11ydx+3ydy=0
anonymous
  • anonymous
I wish that 22 wasn't there. You sure it's not 11?
anonymous
  • anonymous
Positive :/
anonymous
  • anonymous
I know what the answer if, if that helps.
anonymous
  • anonymous
yeah punch it in
anonymous
  • anonymous
Its \[11xy^2+y^3=C\]
anonymous
  • anonymous
Sorry, I keep being called away.
anonymous
  • anonymous
It's alright..
anonymous
  • anonymous
I'm thinking a substitution method. Just one sec...again called away...sorry. Won't be forever.
anonymous
  • anonymous
I think I may have your method...just let me check properly.
anonymous
  • anonymous
Ok cool
anonymous
  • anonymous
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.
anonymous
  • anonymous
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,\[y'=F(\frac{y}{x})\]
anonymous
  • anonymous
We make the identification,\[v(x)=\frac{y}{x}\]so that\[y=xv\]
anonymous
  • anonymous
and we take it from there. Is this familiar?
anonymous
  • anonymous
Ok let me try to pmess with it
anonymous
  • anonymous
Ok instead of that one... can you help me solve y'-9y=8e^x
anonymous
  • anonymous
I'm doing this one for a timed quiz
anonymous
  • anonymous
I figured out your other problem - I made an arithmetic error, but the method is correct.
anonymous
  • anonymous
Yeah I'll do the second.
anonymous
  • anonymous
Ok I think I saw where you were going with it and I gutted my way through it. Thanks
anonymous
  • anonymous
Okay, integrating factor one - do you know?
anonymous
  • anonymous
Not sure what that means..
anonymous
  • anonymous
\[y'-9y=8e^x \rightarrow \frac{d (\mu y)}{dx}=\mu 8e^x\]
anonymous
  • anonymous
\[\mu = e^{\int\limits_{}{}-9dx}=e^{-9x}\]
anonymous
  • anonymous
so
anonymous
  • anonymous
im sorry that should be y"
anonymous
  • anonymous
\[e^{-9x}y=e^{-9x}.8e^x+c \rightarrow y=ce^{9x}-e^x\]
anonymous
  • anonymous
y''-9y=8e^x ?
anonymous
  • anonymous
Yes, sorry for the confusion. I miss typed.
anonymous
  • anonymous
right do you just want the answer for the quiz and then the method
anonymous
  • anonymous
Well. I want to learn but first and foremost I really need to pass this quiz
anonymous
  • anonymous
I am writing it all down to try to understand though.
anonymous
  • anonymous
\[y=c_1e^{3x}+c_2e^{-3x}-e^{x}\]
anonymous
  • anonymous
I'll explain...
anonymous
  • anonymous
Are you sure the + isn't a - between the first 2 terms?
anonymous
  • anonymous
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
anonymous
  • anonymous
+
anonymous
  • anonymous
In the end, it doesn't matter because it's absorbed into the constant.
anonymous
  • anonymous
The only option I have with a + sign in it is... -2e^3x+e^-3x-e^x
anonymous
  • anonymous
Ok so e^3x=e^-3x-e^x is the same thing?
anonymous
  • anonymous
Oh, do you have initial conditions?
anonymous
  • anonymous
Sorry when x=0
anonymous
  • anonymous
Your equation has it's constants solve for
anonymous
  • anonymous
Oh...you need to give me the whole question ;)
anonymous
  • anonymous
Wait.... y=0 and y'=8 and x=0... I'm sorry I've been up all day doing this
anonymous
  • anonymous
It's second order, so there should be 2 conds.
anonymous
  • anonymous
ok
anonymous
  • anonymous
c_1 = 2 and c_2 = -1
anonymous
  • anonymous
\[y=2e^3x-e^{-3x}-e^x\]
anonymous
  • anonymous
Yep that works!
anonymous
  • anonymous
Good
anonymous
  • anonymous
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 :(
anonymous
  • anonymous
It's because you're having to do it on your own - you're not being shown the ropes. Don't punish yourself.
anonymous
  • anonymous
trying not to. Could you help me with 1 or 2 more? Then I'll leave you alone.
anonymous
  • anonymous
yes
anonymous
  • anonymous
How long do you have to do the quiz?
anonymous
  • anonymous
i panic in tests - you shouldn't have told me, lol
anonymous
  • anonymous
4xydx=(x^2+4)dy and that is the whole problem and not leaving anything out :)
anonymous
  • anonymous
It's separable.
anonymous
  • anonymous
\[4xydx=(x^2+4)dy \rightarrow \frac{4xy}{x^2+4}=\frac{dy}{dx} \rightarrow \frac{4x}{x^2+4}dx=\frac{dy}{y}\]
anonymous
  • anonymous
You can integrate from there.
anonymous
  • anonymous
Ok let me try
anonymous
  • anonymous
\[y=c(x^2+4)^2\]
anonymous
  • anonymous
yep thats what I got
anonymous
  • anonymous
Aesome
anonymous
  • anonymous
or, awesome
anonymous
  • anonymous
Ok I think this should be the last one... I think I got the others actually.
anonymous
  • anonymous
ok
anonymous
  • anonymous
dy+2ydx=e^(-4x)dx
anonymous
  • anonymous
It's an integrating factor one again.
anonymous
  • anonymous
Yeah those are really difficult for me. Seperatable ones are the ones I'm more comfortable with...
anonymous
  • anonymous
\[y=-\frac{1}{2}e^{-4x}+ce^{-2x}\]
anonymous
  • anonymous
I'm just skipping to the solution because of the test...
anonymous
  • anonymous
The closest answer I have that looks like that is e^(4x)+Ce^(2x)
anonymous
  • anonymous
hmmm...that's the solution to the equation you gave me...
anonymous
  • anonymous
let me write it in the formula so it looks better maybe it looks funny on text
anonymous
  • anonymous
\[dy+2ydx=e^{-4x}dx \rightarrow y'+2y=e^{-4x}\]?
anonymous
  • anonymous
\[dy-2ydx=6e^(5x) dx\]
anonymous
  • anonymous
that dan 5x is super script
anonymous
  • anonymous
so you gave me the wrong question before?
anonymous
  • anonymous
Im a failure sorry I'm rushing
anonymous
  • anonymous
\[y(x) = c_1 e^{2 x}+2 e^{5 x}\]
anonymous
  • anonymous
is that one available?
anonymous
  • anonymous
yes just not written that way but yeah
anonymous
  • anonymous
I had actually guesed that one after trying to work it myself. Hurray for small miracles.
anonymous
  • anonymous
so you've passed?
anonymous
  • anonymous
Well... not 100% sure. I think I got the rest of them right.
anonymous
  • anonymous
Do you still want to look at the first one?
anonymous
  • anonymous
For dy+2ydx=e^(-4x)dx I got.... y=-1/2e^(-4x)+ce^(-2x)
anonymous
  • anonymous
that's what I got..?
anonymous
  • anonymous
and thats the only other one I kind of guessed on. I do want to look at those though yes.
anonymous
  • anonymous
No I did that one myself
anonymous
  • anonymous
good
anonymous
  • anonymous
I think its right, not 100% sure though
anonymous
  • anonymous
It's right
anonymous
  • anonymous
if
anonymous
  • anonymous
it's the one I did before...and it is...it's right.
anonymous
  • anonymous
You're fine at this!
anonymous
  • anonymous
Oh ok, sorry I'm scattered. ok I'm submitting it!
anonymous
  • anonymous
What about your first question?
anonymous
  • anonymous
How much time do you have?
anonymous
  • anonymous
5 mins left
anonymous
  • anonymous
the first one was a practice but I do want to know how to do it
anonymous
  • anonymous
ok
anonymous
  • anonymous
How about.... is y'=2y(x-3) seperable
anonymous
  • anonymous
I said yes.
anonymous
  • anonymous
yes, separable
anonymous
  • anonymous
ok and 2xdx+ysinxdy=sinxdx
anonymous
  • anonymous
I got y^2=x-2sinx+C
anonymous
  • anonymous
that's separable too - is that how you solved it?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
that one would have taken a bit of effort
anonymous
  • anonymous
Ok is y'=y/x y=cx
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok well it looks all good then :)
anonymous
  • anonymous
Thank you so much for that man.....
anonymous
  • anonymous
that's okay - i hope you pass
anonymous
  • anonymous
have you heard of a site called wolframalpha? www.wolframalpha.com
anonymous
  • anonymous
Ok so that original problem..
anonymous
  • anonymous
You can type in things and it will calculate, like your diff. eqs
anonymous
  • anonymous
but it won't show you what to do, that's the only thing
anonymous
  • anonymous
I have, I don't know how to type these kinds of problems in though.... could you show me how?
anonymous
  • anonymous
I'm just thinking, if you have more of these tests, you can have a look at that as a resource
anonymous
  • anonymous
I did use it before, I have 2 more tests actually.
anonymous
  • anonymous
y'-2y=e^(-4x) is what you can type in for one of your d.e.'s above. It will give you the answer.
anonymous
  • anonymous
Use primes on the y's...
anonymous
  • anonymous
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.
anonymous
  • anonymous
Hmm.... but even for stuff like... 22xdy+11ydx+3ydy=0?
anonymous
  • anonymous
Yeah, divide everything on your paper by dx so you can have (dy/dx) which you can equate as y'
anonymous
  • anonymous
lol... thats a genius way to do it.
anonymous
  • anonymous
\[22x \frac{dy}{dx}+y+3y \frac{dy}{dx}=0 \rightarrow 22xy'+y+3yy'=0\]
anonymous
  • anonymous
I don't know why they keep giving you the equation in that form...it's weird.
anonymous
  • anonymous
In the form I gave you? Thats the way my teacher gives me them....
anonymous
  • anonymous
Yeah, I know...weird.
anonymous
  • anonymous
Almost all my questions are like that is that not how they are normally done?
anonymous
  • anonymous
lol, no
anonymous
  • anonymous
Hmm why would she try to teach like that then?
anonymous
  • anonymous
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.
anonymous
  • anonymous
did you pass this quiz?
anonymous
  • anonymous
Ok well I really appreciate it. You might be the only reason I passed that quiz... I'm sure of it.
anonymous
  • anonymous
I won't know for a day or two, but I'm sure I did.
anonymous
  • anonymous
Cool, hope so...don't want to be spending more money on tuition...
anonymous
  • anonymous
god I know. I have...2 more quizes and a final. Then I'll be done with this class.
anonymous
  • anonymous
Good luck then. If you need help, look up an old post and send something. I'll get an e-mail about it.
anonymous
  • anonymous
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
anonymous
  • anonymous
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 ;)
anonymous
  • anonymous
I'm going to go offline for a while. I have some stuff to do. Good luck!

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