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Astrophysics
 one year ago
@ganeshie8
Astrophysics
 one year ago
@ganeshie8

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1How can I tell where this solution is valid, the \[(2xy)dx+(2yx)dy = 0\] solving this exact equation I got \[\psi (x,y) = x^2xy+y^27\] but I'm not sure how to tell where this is valid

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what are the initial conditions ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh oops! y(1) = 3

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0are you taking a DE COURSE???

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so the solution curve is \(x^2xy+y^27=0\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1the back of the book says \[x<\sqrt{\frac{ 28 }{ 3 }}\]

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0STUDY WITH ME IM LIKE 5 weeks behind LOL in class we're like up to PDE's and im still stuggling with fourier :/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443339556634:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Notice that the solution curve is trapped between those vertical tangents we may use that observation to find the `x` bounds

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yup I see, but I'm not sure how to do it algebraically

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2As a start, implicitly differentiate the equation of solution curve and find an expression for \(y'\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2alternatively 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
 one year ago
Best ResponseYou've already chosen the best response.1\[y' = \frac{ 2xy }{ x2y }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ooh so how would you set that up using quadratic formula

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2in 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 \(x\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2If that derivative thing doesn't strike immediately, you may simply use the disrcriminat : \(\color{red}{y^2}\color{red}{y}x+x^27=0\) this is a quadratic in \(\color{red}{y}\) : \(a=1\) \(b=x\) \(c=x^27\) Discriminant must be nonnegative for \(y\) to be real. so \(x^24(1)(x^27)\ge 0\), we can solve \(x\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ha, I like the derivative thing, so this is finding the same thing as interval of definition?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes this is like finding the largest interval in which the solution curve is defined

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So how come it doesn't say \[\sqrt{\frac{ 28 }{ 3 }} < x < \sqrt{\frac{ 28 }{ 3 }}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1< = to I believe it should be

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dy/dx is not defined at either ends, so we must exclude them

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the derivative thingy is easy because the derivative is readily available for us : \((2xy)dx+(2yx)dy = 0 \implies \dfrac{dy}{dx}=\dfrac{y2x}{2yx}\) Clearly you don't wanto plug \(2y=x\) in above equation

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yup, I really like this method

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Solving 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 >.<

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the interval stuff is really not that easy, there are several things that you need to consider when working the domain of the solution

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2this 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
 one year ago
Best ResponseYou've already chosen the best response.1Awesome, 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"

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ah ok haha, thanks xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2except for philosophical differences, both refer to the same thing

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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
 one year ago
Best ResponseYou've already chosen the best response.21) "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
 one year ago
Best ResponseYou've already chosen the best response.1Ok that makes sense, like I had a separable equation problem earlier and it asked for where it was defined \[\frac{ dy }{ dx } = \frac{ arcsinx }{ y^2(1x^2)^{1/2} }\] I sort of just looked at the derivative here and got 1<x<1, but I guess I should be paying attention to vertical tangents as well?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what are the initial conditions ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2whats the solution ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you must consider the DE, the initial conditions and the solution while working the domain

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2All 3 things can change the domain

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok yeah that's what I did, that was sort of a bad example

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I didn't include all the info

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1The solution was \[y(x) = \left[ \frac{ 3 }{ 2 }\arcsin^2x \right]^{1/3}+1\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[y^2dy = \frac{ arcsinx }{ (1x^2) }dx\] y(0)=1 so not too hard of a problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.21<x<1 looks good to me why do you think it is wrong ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh no I didn't think it was wrong it made sense to me, but I didn't really consider the vertical tangent

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok nvm, I think I have got a decent idea of this stuff now hehe, thanks!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2try finding the domain of this IVP \[\dfrac{dy}{dx}=\dfrac{2}{\sqrt{y}};~~y(0) = 9\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Omg C seems ugly X_X

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.1Ok haha, \[y=3^{2/3}(x+9)^{2/3}\] so the domain then is x>9

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what happens if i plugin x = 10

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so nothing wrong with x<9 ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Nope, so all real numbers then?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Look at the graph of solution again : dw:1443344660260:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2then look at the differential equation again

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the graph is not matching with the given differential equation, why ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443344789571:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Nooo, because we then get undefined

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right, the slope of tangent to solution curve, y, is negative for y < 9, yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443345018801:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1But see that's where I was thinking, when we look at the derivative itself we would have to consider the vertical tangets

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2its 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
 one year ago
Best ResponseYou've already chosen the best response.2so we must exclude x<9 from the domain of the solutions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2as you can see, finding domain is not so straightforward and definitely not easy

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, it's pretty confusing haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So we're sort of restricting it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2since the derivative is always "positive", x <= 9 makes no sense, so the domain is x > 9

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh haha I'm reading the link, that's where you got the problem

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1This stuff is confusing haha it's like I get it and at the same time I don't

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2domains 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
 one year ago
Best ResponseYou've already chosen the best response.1I guess, I'll have to do lots of practice

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Well thanks a lot ganeshie it's nearly 4 am, been doing this stuff nearly all day haha, time for sleep, take care, bye!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2tc, have good sleep :)
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