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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
what are the initial conditions ?
Oh oops! y(1) = 3

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are you taking a DE COURSE???
study with me pls <3
so the solution curve is \(x^2-xy+y^2-7=0\)
Haha yeah I am
Yes
the back of the book says \[|x|<\sqrt{\frac{ 28 }{ 3 }}\]
STUDY WITH ME IM LIKE 5 weeks behind LOL in class we're like up to PDE's and im still stuggling with fourier :/
|dw:1443339556634:dw|
Notice that the solution curve is trapped between those vertical tangents we may use that observation to find the `x` bounds
Yup I see, but I'm not sure how to do it algebraically
As a start, implicitly differentiate the equation of solution curve and find an expression for \(y'\)
kk
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
\[y' = \frac{ 2x-y }{ x-2y }\]
Ooh so how would you set that up using quadratic formula
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 \(x\)
That works!
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 \(x\)
Ha, I like the derivative thing, so this is finding the same thing as interval of definition?
Yes this is like finding the largest interval in which the solution curve is defined
So how come it doesn't say \[-\sqrt{\frac{ 28 }{ 3 }} < x < \sqrt{\frac{ 28 }{ 3 }}\]
< = to I believe it should be
dy/dx is not defined at either ends, so we must exclude them
Oh right!
Haha
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
Yup, I really like this method
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 >.<
Thanks again
the interval stuff is really not that easy, there are several things that you need to consider when working the domain of the solution
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
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"
both are same
Ah ok haha, thanks xD
except for philosophical differences, both refer to the same thing
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
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..
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
what are the initial conditions ?
y(0)=1
whats the solution ?
.
you must consider the DE, the initial conditions and the solution while working the domain
All 3 things can change the domain
Ok yeah that's what I did, that was sort of a bad example
I didn't include all the info
The solution was \[y(x) = \left[ \frac{ 3 }{ 2 }\arcsin^2x \right]^{1/3}+1\]
\[y^2dy = \frac{ arcsinx }{ (1-x^2) }dx\] y(0)=1 so not too hard of a problem
-1
Oh no I didn't think it was wrong it made sense to me, but I didn't really consider the vertical tangent
Ok nvm, I think I have got a decent idea of this stuff now hehe, thanks!
try finding the domain of this IVP \[\dfrac{dy}{dx}=\dfrac{2}{\sqrt{y}};~~y(0) = 9\]
Ok give me a sec
Omg C seems ugly X_X
nvm its 18?
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
Ok haha, \[y=3^{2/3}(x+9)^{2/3}\] so the domain then is x>-9
\[x \ge -9\]
why
what happens if i plugin x = -10
you get -3^(2/3)
So that works
so nothing wrong with x<-9 ?
Nope, so all real numbers then?
Look at the graph of solution again : |dw:1443344660260:dw|
then look at the differential equation again
the graph is not matching with the given differential equation, why ?
|dw:1443344789571:dw|
Nooo, because we then get undefined
right, the slope of tangent to solution curve, y, is negative for y < -9, yes ?
Yup
|dw:1443345018801:dw|
But see that's where I was thinking, when we look at the derivative itself we would have to consider the vertical tangets
tangets*
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
so we must exclude x<-9 from the domain of the solutions
Ah!
as you can see, finding domain is not so straightforward and definitely not easy
Yeah, it's pretty confusing haha
So we're sort of restricting it?
since the derivative is always "positive", x <= -9 makes no sense, so the domain is x > -9
Ok I see
Oh haha I'm reading the link, that's where you got the problem
This stuff is confusing haha it's like I get it and at the same time I don't
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...
I guess, I'll have to do lots of practice
Well thanks a lot ganeshie it's nearly 4 am, been doing this stuff nearly all day haha, time for sleep, take care, bye!
tc, have good sleep :)

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