Question

@ganeshie8

Answer 1

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

Answer 2

what are the initial conditions ?

Answer 3

Oh oops! y(1) = 3

Answer 4

are you taking a DE COURSE???

Answer 5

study with me pls <3

Answer 6

so the solution curve is \(x^2-xy+y^2-7=0\)

Answer 7

Haha yeah I am

Answer 8

Yes

Answer 9

the back of the book says \[|x|<\sqrt{\frac{ 28 }{ 3 }}\]

Answer 10

STUDY WITH ME IM LIKE 5 weeks behind LOL
in class we're like up to PDE's
and im still stuggling with fourier :/

Answer 11

Notice that the solution curve is trapped between those vertical tangents

we may use that observation to find the `x` bounds

Answer 12

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

Answer 13

As a start, implicitly differentiate the equation of solution curve and find an expression for \(y'\)

Answer 14

kk

Answer 15

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

Answer 16

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

Answer 17

Ooh so how would you set that up using quadratic formula

Answer 18

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 \(x\)

Answer 19

That works!

Answer 20

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 \(x\)

Answer 21

Ha, I like the derivative thing, so this is finding the same thing as interval of definition?

Answer 22

Yes this is like finding the largest interval in which the solution curve is defined

Answer 23

So how come it doesn't say \[-\sqrt{\frac{ 28 }{ 3 }} < x < \sqrt{\frac{ 28 }{ 3 }}\]

Answer 24

< = to I believe it should be

Answer 25

dy/dx is not defined at either ends, so we must exclude them

Answer 26

Oh right!

Answer 27

Haha

Answer 28

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

Answer 29

Yup, I really like this method

Answer 30

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 >.<

Answer 31

Thanks again

Answer 32

the interval stuff is really not that easy, there are several things that you need to consider when working the domain of the solution

Answer 33

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

Answer 34

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"

Answer 35

both are same

Answer 36

Ah ok haha, thanks xD

Answer 37

except for philosophical differences, both refer to the same thing

Answer 38

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

Answer 39

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..

Answer 40

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<x<1, but I guess I should be paying attention to vertical tangents as well?

Answer 41

what are the initial conditions ?

Answer 42

y(0)=1

Answer 43

whats the solution ?

Answer 44

.

Answer 45

you must consider the DE, the initial conditions and the solution while working the domain

Answer 46

All 3 things can change the domain

Answer 47

Ok yeah that's what I did, that was sort of a bad example

Answer 48

I didn't include all the info

Answer 49

The solution was \[y(x) = \left[ \frac{ 3 }{ 2 }\arcsin^2x \right]^{1/3}+1\]

Answer 50

\[y^2dy = \frac{ arcsinx }{ (1-x^2) }dx\] y(0)=1 so not too hard of a problem

Answer 51

-1<x<1 looks good to me
why do you think it is wrong ?

Answer 52

Oh no I didn't think it was wrong it made sense to me, but I didn't really consider the vertical tangent

Answer 53

Ok nvm, I think I have got a decent idea of this stuff now hehe, thanks!

Answer 54

try finding the domain of this IVP

\[\dfrac{dy}{dx}=\dfrac{2}{\sqrt{y}};~~y(0) = 9\]

Answer 55

Ok give me a sec

Answer 56

Omg C seems ugly X_X

Answer 57

nvm its 18?

Answer 58

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

Answer 59

Ok haha, \[y=3^{2/3}(x+9)^{2/3}\] so the domain then is x>-9

Answer 60

\[x \ge -9\]

Answer 61

why

Answer 62

what happens if i plugin x = -10

Answer 63

you get -3^(2/3)

Answer 64

So that works

Answer 65

so nothing wrong with x<-9 ?

Answer 66

Nope, so all real numbers then?

Answer 67

Look at the graph of solution again :

Answer 68

then look at the differential equation again

Answer 69

the graph is not matching with the given differential equation, why ?