Differential Equations Watch

vector
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#1
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Just trying to get my head around differential equations but still not really sure about them I'm trying to do

xy'+y+4=0 and y'+y=xy^{\frac{2}{3}}

From a bit of research I'm thinking that the 1st is homogeneous and the 2nd is linear? Is that right?
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Kolya
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The first is not homogeneous, and the second is not linear. :no: A DE is linear if it can be written in the form a_n(x) y^{[n]} + ... + a_1(x) y' + a_0(x) y = f(x) , and is homogeneous if it is linear and f(x) = 0 (in the form I have written out). So the first is inhomogeneous because f(x) = -4, and the second is nonlinear because we have a xy^{2/3} term.
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DFranklin
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Be careful: homogeneous is a slippery term:
(Original post by Wiki)
A homogeneous differential equation has several distinct meanings.
See http://en.wikipedia.org/wiki/Homogen...ntial_equation
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vector
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(Original post by Kolya)
The first is not homogeneous, and the second is not linear. :no: A DE is linear if it can be written in the form , and is homogeneous if it is linear and f(x) = 0 (in the form I have written out). So the first is inhomogeneous because f(x) = -4, and the second is nonlinear because we have a term.
Fantastic All our notes have is methods for solving separable, homogeneous, linear and Bernouilli, there's no clue as to how to distinguish between them Which method would I be looking at for these two then?
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harr
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Try inspection for the first one. What differentiates to give xy'+y?

Edit: The first equation is homogeneous, so if you can't do it by inspection, the standard methods you've been given should work. If you can't do it from your notes, just google a few examples to see how to do things in that general form. It's probably worth doing this one in both ways for practice.
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vector
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(Original post by harr)
Try inspection for the first one. What differentiates to give xy'+y?

Edit: The first equation is homogeneous, so if you can't do it by inspection, the standard methods you've been given should work. If you can't do it from your notes, just google a few examples to see how to do things in that general form. It's probably worth doing this one in both ways for practice.
Ah, ok, xy differentiates to give that, but how does that help?
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Kolya
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Write it as d/dx (xy) = -4 , and integrate both sides w.r.t. x
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DFranklin
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(Original post by vector)
Ah, ok, xy differentiates to give that, but how does that help?
If \frac{d}{dx} f(x,y) = g(x), then integrating both sides w.r.t. x f(x,y) = \int g(x)\,dx

For the 2nd one, I think the substitution y=z^3 brings you to something more tractable.
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Kolya
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If for some reason you don't spot the rearrangement of the LHS for the 1st one, then the standard method still works. You can rewrite the equation in the form y' + Q(x)y = P(x) , and then use an "integrating factor", I(x) = e^{\int Q(x) \ dx}. This gives you the equation you started with (surprise, surprise), and tells you that d/dx (I(x)y) = d/dx (xy). It's much easier to just spot the form of the LHS, though!
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