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what are the initial conditions ?

Oh oops! y(1) = 3

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 \[|x|<\sqrt{\frac{ 28 }{ 3 }}\]

|dw:1443339556634:dw|

Yup I see, but I'm not sure how to do it algebraically

kk

\[y' = \frac{ 2x-y }{ x-2y }\]

Ooh so how would you set that up using quadratic formula

That works!

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

Yup, I really like this method

Thanks again

both are same

Ah ok haha, thanks xD

except for philosophical differences, both refer to the same thing

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

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?

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|

tangets*

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

I guess, I'll have to do lots of practice

tc, have good sleep :)