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

- anonymous

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

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- anonymous

22xdy+11ydx+3ydy=0

- anonymous

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

- anonymous

Hehe, I see...

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## More answers

- anonymous

Are you studying exact differential equations?

- anonymous

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

- anonymous

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

- anonymous

Sure

- anonymous

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

- anonymous

ok

- anonymous

i need paper

- anonymous

haha alright

- anonymous

oh, i don't have her question.

- anonymous

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

- anonymous

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

- 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

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

- anonymous

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

- anonymous

Positive :/

- anonymous

I know what the answer if, if that helps.

- anonymous

yeah punch it in

- anonymous

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

- anonymous

Sorry, I keep being called away.

- anonymous

It's alright..

- anonymous

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

- anonymous

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

- anonymous

Ok cool

- 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

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

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

- anonymous

and we take it from there. Is this familiar?

- anonymous

Ok let me try to pmess with it

- anonymous

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

- anonymous

I'm doing this one for a timed quiz

- anonymous

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

- anonymous

Yeah I'll do the second.

- anonymous

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

- anonymous

Okay, integrating factor one - do you know?

- anonymous

Not sure what that means..

- anonymous

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

- anonymous

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

- anonymous

so

- anonymous

im sorry that should be
y"

- anonymous

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

- anonymous

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

- anonymous

Yes, sorry for the confusion. I miss typed.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I'll explain...

- anonymous

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

- 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

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

- anonymous

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

- anonymous

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

- anonymous

Oh, do you have initial conditions?

- anonymous

Sorry when x=0

- anonymous

Your equation has it's constants solve for

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

c_1 = 2 and c_2 = -1

- anonymous

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

- anonymous

Yep that works!

- anonymous

Good

- 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

It's because you're having to do it on your own - you're not being shown the ropes. Don't punish yourself.

- anonymous

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

- anonymous

yes

- anonymous

How long do you have to do the quiz?

- anonymous

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

- anonymous

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

- anonymous

It's separable.

- 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

You can integrate from there.

- anonymous

Ok let me try

- anonymous

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

- anonymous

yep thats what I got

- anonymous

Aesome

- anonymous

or, awesome

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

It's an integrating factor one again.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

that dan
5x is super script

- anonymous

so you gave me the wrong question before?

- anonymous

Im a failure sorry I'm rushing

- anonymous

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

- anonymous

is that one available?

- anonymous

yes just not written that way but yeah

- anonymous

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

- anonymous

so you've passed?

- anonymous

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

- anonymous

Do you still want to look at the first one?

- anonymous

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

- anonymous

that's what I got..?

- anonymous

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

- anonymous

No I did that one myself

- anonymous

good

- anonymous

I think its right, not 100% sure though

- anonymous

It's right

- anonymous

if

- anonymous

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

- anonymous

You're fine at this!

- anonymous

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

- anonymous

What about your first question?

- anonymous

How much time do you have?

- anonymous

5 mins left

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

I said yes.

- anonymous

yes, separable

- anonymous

ok and 2xdx+ysinxdy=sinxdx

- anonymous

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

- anonymous

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

- anonymous

Yes

- anonymous

that one would have taken a bit of effort

- anonymous

Ok is y'=y/x y=cx

- anonymous

yes

- anonymous

ok well it looks all good then :)

- anonymous

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

- anonymous

that's okay - i hope you pass

- anonymous

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

- anonymous

Ok so that original problem..

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

Use primes on the y's...

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Yeah, I know...weird.

- anonymous

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

- anonymous

lol, no

- anonymous

Hmm why would she try to teach like that then?

- 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

did you pass this quiz?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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

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

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

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

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