Astrophysics
  • Astrophysics
@ganeshie8
Mathematics
  • Stacey Warren - Expert brainly.com
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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!
Astrophysics
  • Astrophysics
How can I tell where this solution is valid, the \[(2x-y)dx+(2y-x)dy = 0\] solving this exact equation I got \[\psi (x,y) = x^2-xy+y^2-7\] but I'm not sure how to tell where this is valid
ganeshie8
  • ganeshie8
what are the initial conditions ?
Astrophysics
  • Astrophysics
Oh oops! y(1) = 3

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Mimi_x3
  • Mimi_x3
are you taking a DE COURSE???
Mimi_x3
  • Mimi_x3
study with me pls <3
ganeshie8
  • ganeshie8
so the solution curve is \(x^2-xy+y^2-7=0\)
Astrophysics
  • Astrophysics
Haha yeah I am
Astrophysics
  • Astrophysics
Yes
Astrophysics
  • Astrophysics
the back of the book says \[|x|<\sqrt{\frac{ 28 }{ 3 }}\]
Mimi_x3
  • Mimi_x3
STUDY WITH ME IM LIKE 5 weeks behind LOL in class we're like up to PDE's and im still stuggling with fourier :/
ganeshie8
  • ganeshie8
|dw:1443339556634:dw|
ganeshie8
  • ganeshie8
Notice that the solution curve is trapped between those vertical tangents we may use that observation to find the `x` bounds
Astrophysics
  • Astrophysics
Yup I see, but I'm not sure how to do it algebraically
ganeshie8
  • ganeshie8
As a start, implicitly differentiate the equation of solution curve and find an expression for \(y'\)
Astrophysics
  • Astrophysics
kk
ganeshie8
  • ganeshie8
alternatively you may think of the solution curve as a quadratic in \(y\), then use the discriminant to see for what values of \(x\), the value of \(y\) is real
Astrophysics
  • Astrophysics
\[y' = \frac{ 2x-y }{ x-2y }\]
Astrophysics
  • Astrophysics
Ooh so how would you set that up using quadratic formula
ganeshie8
  • ganeshie8
in above expression for derivative, \(x=2y\) makes the derivative go wild, so vertical tangents must occur when \(x=2y\) using that eliminate \(y\) in the solution curve and solve \(x\)
Astrophysics
  • Astrophysics
That works!
ganeshie8
  • ganeshie8
If that derivative thing doesn't strike immediately, you may simply use the disrcriminat : \(\color{red}{y^2}-\color{red}{y}x+x^2-7=0\) this is a quadratic in \(\color{red}{y}\) : \(a=1\) \(b=-x\) \(c=x^2-7\) Discriminant must be nonnegative for \(y\) to be real. so \(x^2-4(1)(x^2-7)\ge 0\), we can solve \(x\)
Astrophysics
  • Astrophysics
Ha, I like the derivative thing, so this is finding the same thing as interval of definition?
ganeshie8
  • ganeshie8
Yes this is like finding the largest interval in which the solution curve is defined
Astrophysics
  • Astrophysics
So how come it doesn't say \[-\sqrt{\frac{ 28 }{ 3 }} < x < \sqrt{\frac{ 28 }{ 3 }}\]
Astrophysics
  • Astrophysics
< = to I believe it should be
ganeshie8
  • ganeshie8
dy/dx is not defined at either ends, so we must exclude them
Astrophysics
  • Astrophysics
Oh right!
Astrophysics
  • Astrophysics
Haha
ganeshie8
  • ganeshie8
the derivative thingy is easy because the derivative is readily available for us : \((2x-y)dx+(2y-x)dy = 0 \implies \dfrac{dy}{dx}=\dfrac{y-2x}{2y-x}\) Clearly you don't wanto plug \(2y=x\) in above equation
Astrophysics
  • Astrophysics
Yup, I really like this method
Astrophysics
  • Astrophysics
Solving O.D.E's isn't too bad, it's this interval stuff that can be confusing as it requires you to remember stuff from calc 1 >.<
Astrophysics
  • Astrophysics
Thanks again
ganeshie8
  • ganeshie8
the interval stuff is really not that easy, there are several things that you need to consider when working the domain of the solution
ganeshie8
  • ganeshie8
this pdf talks about "why" domain is important and "how" to find the domain http://apcentral.collegeboard.com/apc/members/repository/ap07_calculus_DE_domain_fin.pdf
Astrophysics
  • Astrophysics
Awesome, thanks by the way...what exactly is the difference from the problems that say "determine the interval in which the solution is defined" and from this where it asks you for "where solution is valid"
ganeshie8
  • ganeshie8
both are same
Astrophysics
  • Astrophysics
Ah ok haha, thanks xD
ganeshie8
  • ganeshie8
except for philosophical differences, both refer to the same thing
Astrophysics
  • Astrophysics
Ok cool, yeah I guess it's just they word each question differently, so I thought there must be something different about them even though they sound the same
ganeshie8
  • ganeshie8
1) "determine the interval in which the solution is defined" 2) "where solution is valid" first one is in ur control, "You" can define the interval where your solution would be valid while modeling the differential equaiton based on physical constraints. for the second one, you need to work out the domain by looking at the solution but there is not much difference between them, at least i never worried about the differences..
Astrophysics
  • Astrophysics
Ok that makes sense, like I had a separable equation problem earlier and it asked for where it was defined \[\frac{ dy }{ dx } = \frac{ arcsinx }{ y^2(1-x^2)^{1/2} }\] I sort of just looked at the derivative here and got -1
ganeshie8
  • ganeshie8
what are the initial conditions ?
Astrophysics
  • Astrophysics
y(0)=1
ganeshie8
  • ganeshie8
whats the solution ?
IrishBoy123
  • IrishBoy123
.
ganeshie8
  • ganeshie8
you must consider the DE, the initial conditions and the solution while working the domain
ganeshie8
  • ganeshie8
All 3 things can change the domain
Astrophysics
  • Astrophysics
Ok yeah that's what I did, that was sort of a bad example
Astrophysics
  • Astrophysics
I didn't include all the info
Astrophysics
  • Astrophysics
The solution was \[y(x) = \left[ \frac{ 3 }{ 2 }\arcsin^2x \right]^{1/3}+1\]
Astrophysics
  • Astrophysics
\[y^2dy = \frac{ arcsinx }{ (1-x^2) }dx\] y(0)=1 so not too hard of a problem
ganeshie8
  • ganeshie8
-1
Astrophysics
  • Astrophysics
Oh no I didn't think it was wrong it made sense to me, but I didn't really consider the vertical tangent
Astrophysics
  • Astrophysics
Ok nvm, I think I have got a decent idea of this stuff now hehe, thanks!
ganeshie8
  • ganeshie8
try finding the domain of this IVP \[\dfrac{dy}{dx}=\dfrac{2}{\sqrt{y}};~~y(0) = 9\]
Astrophysics
  • Astrophysics
Ok give me a sec
Astrophysics
  • Astrophysics
Omg C seems ugly X_X
Astrophysics
  • Astrophysics
nvm its 18?
ganeshie8
  • ganeshie8
you may simply use wolfram to find the solution http://www.wolframalpha.com/input/?i=solve+dy%2Fdx%3D2%2Fsqrt%28y%29%2C+y%280%29%3D9
Astrophysics
  • Astrophysics
Ok haha, \[y=3^{2/3}(x+9)^{2/3}\] so the domain then is x>-9
Astrophysics
  • Astrophysics
\[x \ge -9\]
ganeshie8
  • ganeshie8
why
ganeshie8
  • ganeshie8
what happens if i plugin x = -10
Astrophysics
  • Astrophysics
you get -3^(2/3)
Astrophysics
  • Astrophysics
So that works
ganeshie8
  • ganeshie8
so nothing wrong with x<-9 ?
Astrophysics
  • Astrophysics
Nope, so all real numbers then?
ganeshie8
  • ganeshie8
Look at the graph of solution again : |dw:1443344660260:dw|
ganeshie8
  • ganeshie8
then look at the differential equation again
ganeshie8
  • ganeshie8
the graph is not matching with the given differential equation, why ?
ganeshie8
  • ganeshie8
|dw:1443344789571:dw|
Astrophysics
  • Astrophysics
Nooo, because we then get undefined
ganeshie8
  • ganeshie8
right, the slope of tangent to solution curve, y, is negative for y < -9, yes ?
Astrophysics
  • Astrophysics
Yup
ganeshie8
  • ganeshie8
|dw:1443345018801:dw|
Astrophysics
  • Astrophysics
But see that's where I was thinking, when we look at the derivative itself we would have to consider the vertical tangets
Astrophysics
  • Astrophysics
tangets*
ganeshie8
  • ganeshie8
its not primarily about vertical tangents here its much more deeper than that, the slope field of the given differential equation cannot have negative slope segments
ganeshie8
  • ganeshie8
so we must exclude x<-9 from the domain of the solutions
Astrophysics
  • Astrophysics
Ah!
ganeshie8
  • ganeshie8
as you can see, finding domain is not so straightforward and definitely not easy
Astrophysics
  • Astrophysics
Yeah, it's pretty confusing haha
Astrophysics
  • Astrophysics
So we're sort of restricting it?
ganeshie8
  • ganeshie8
since the derivative is always "positive", x <= -9 makes no sense, so the domain is x > -9
Astrophysics
  • Astrophysics
Ok I see
Astrophysics
  • Astrophysics
Oh haha I'm reading the link, that's where you got the problem
Astrophysics
  • Astrophysics
This stuff is confusing haha it's like I get it and at the same time I don't
ganeshie8
  • ganeshie8
domains for differential equations are not same as that of algebraic equations nothing to feel bad about, it takes a while to get used to...
Astrophysics
  • Astrophysics
I guess, I'll have to do lots of practice
Astrophysics
  • Astrophysics
Well thanks a lot ganeshie it's nearly 4 am, been doing this stuff nearly all day haha, time for sleep, take care, bye!
ganeshie8
  • ganeshie8
tc, have good sleep :)

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