Differential equations Watch

The Muon
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#1
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#1
Struggling a bit with this. not quite sure how to turn the \frac{dy}{dx} into a \frac{d}{dx}

Question
Write the left hand side of the differential equation

(2x+y)+(x+2y)\dfrac{dy}{dx} = 0

in the form

\dfrac{d}{dx}(F(x,y))


The question does go on to get you to solve it but i think I may be able to solve then when I have it. If not there will undoubtedly be more posts :tongue:
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generalebriety
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#2
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You need to look for things that, when differentiated (using the product rule), give you terms looking like x.dy/dx and 2y.dy/dx. The latter should be obvious, the former should be easy to guess. After you've done that, the rest of the terms should fall into place.
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around
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#3
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Looks like an application of the product rule.

don't click here
have a look at (x^2 + xy+ y^2)


edit: corrected horrible mistakes
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The Muon
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yeah - got it now - it was the product rule but I was missing :facepalm:.
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The Muon
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#5
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I lied - i got the F(x,y) but im stuck with solving it.

\dfrac{d}{dx} \big (x^2 + xy + y^2) = 0

is what i now have but I don't think I know how to solve an equation of this form :confused:
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around
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#6
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I'm missing something really obvious but I'd just say integrate both sides wrt x...

(if you could write it explicitly then there would be no need for implicit shenanigans)
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generalebriety
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(Original post by The Muon)
\dfrac{d}{dx} \big (\underline{\underline{x^2 + xy + y^2}}) = 0
What functions have zero derivative? So what type of function must this be?
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Mathletics
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(Original post by The Muon)
I lied - i got the F(x,y) but im stuck with solving it.

\dfrac{d}{dx} \big (x^2 + xy + y^2) = 0

is what i now have but I don't think I know how to solve an equation of this form :confused:
Can't you just seperate the variables and integrate.

\displaystyle\int d(x^2 + xy + y^2) = \displaystyle\int 0\ dx

x^2 + xy + y^2 = c
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The Muon
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I wouldn't know how to integrate y with respect to x

And would a straight line have a zero derivative?

I don't even know what I'm looking for :confused: should I come out with a y= or an x= or something similar to what is in the brackets?

Its probably easier if you have any resources or can point me to a webpage because this is all new to me :rolleyes:
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Swayum
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(Original post by The Muon)
I wouldn't know how to integrate y with respect to x
You're not being asked to. Let z = x^2 + xy + y^2. Then you have

dz/dx = 0

Which I'm sure you'll agree integrates to z = c, yes? i.e. x^2 + xy + y^2 = c. Done.

And would a straight line have a zero derivative?
:lolwut:

y = 2x - 5 is a straight line with a non-zero derivative (as we learned in Year 8 :p:). Yup, this is the part you facepalm yourself.
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Mathletics
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(Original post by The Muon)
I wouldn't know how to integrate y with respect to x

And would a straight line have a zero derivative?

I don't even know what I'm looking for :confused: should I come out with a y= or an x= or something similar to what is in the brackets?

Its probably easier if you have any resources or can point me to a webpage because this is all new to me :rolleyes:
You're not integrating y with respect to x, you're integrating 1 with respect to x^2 + xy + y^2.

Also, what happens when you differentiate a constant?
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generalebriety
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(Original post by The Muon)
I wouldn't know how to integrate y with respect to x
But you're not trying to. You're integrating a derivative.

My point was that only constants have derivative equal to zero, so that mess you have whose derivative is zero must be constant - i.e. x^2 + xy + y^2 = k. Alternatively, put z = x^2 + xy + y^2, and then dz/dx = 0, so z = k. Or do it the way Mathletics did it, which was the same, except he didn't call it z first.

Integration and differentiation are (under certain conditions, almost) inverse operations. Look up the 'fundamental theorem of calculus'.
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The Muon
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#13
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(Original post by Swayum)
You're not being asked to. Let z = x^2 + xy + y^2. Then you have

dz/dx = 0

Which I'm sure you'll agree integrates to z = c, yes? i.e. x^2 + xy + y^2 = c. Done.



:lolwut:

y = 2x - 5 is a straight line with a non-zero derivative (as we learned in Year 8 :p:). Yup, this is the part you facepalm yourself.
yeah - i got it now, A facepalm is truly in order. I was imagining a horizontal line which is straight I suppose :facepalm: :facepalm2:

Thankees guys.
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around
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You don't need to worry about integrating y wrt x; integration is the inverse of differentiation and hence \int \frac{d}{dx} F(x,y) dx= F(x,y)

EDIT: wouldn't the resulting function just be a horizontal line, given that d/dx is uniformly 0?
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DFranklin
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#15
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You can use the quadratic formula to find y as a function of x if you really want.
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The Muon
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#16
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might as well keep my questions in the same thread -keep things tidy and whatnot.
I have to now use the substitution y(x)=xv(x). Does this simply mean replace y with xv and \frac{dy}{dx} with v+ x\frac{dv}{dx}?
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Oh I Really Don't Care
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Yes
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The Muon
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#18
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(Original post by DeanK22)
Yes
clear and succinct - I like it :yep:
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The Muon
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#19
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#19
Another one

If i make the substitutions

x=X+a \newline

y=Y+b

what happens to my \frac{dy}{dx}

I'm tempted it stays the same as the a and b are just constants but I would like some confirmation before I spend a while doing the question.

Cheers
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The Muon
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#20
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(Original post by nuodai)
Well if you differentiate the Y equation with respect to x you get \frac{dy}{dx} = \frac{dY}{dx}, which is equal to \frac{dY}{dx} \frac{dx}{dX}... can you see where to go now?
I don't think I understand how the two are equal - unless im missing something stupidly obvious :confused:

Edit: does it turn from \frac{dy}{dx} \text{to} \frac{dY}{dX}
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