bahrom7893
  • bahrom7893
Differential equations help! y*dx + (3+3x-y)*dy = 0 The topic is integrating factors The answer is 4xy^3 + 4y^3 - y^4 = C
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
hi bahrom
anonymous
  • anonymous
want me to use that list thingy
bahrom7893
  • bahrom7893
u dont have to

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bahrom7893
  • bahrom7893
just a solution, its integrating factors, its just my brains are dead
anonymous
  • anonymous
what do you mean by integrating factors?
bahrom7893
  • bahrom7893
you know when you let something be [e^Integral(f(x;y))dx] then multiply across by it to get a derivative of some expression on the right and some ez integral on the left or vice versa
anonymous
  • anonymous
oh
anonymous
  • anonymous
so you have y + dy/dx ( 3 + 3x -y) = 0
bahrom7893
  • bahrom7893
well yeah then move y over
sasogeek
  • sasogeek
does it mean you're basically trying to prove the answer?
bahrom7893
  • bahrom7893
no I just know the answer
anonymous
  • anonymous
well for integrating factors we want dy/dx + p(x)y = g(x)
anonymous
  • anonymous
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
bahrom7893
  • bahrom7893
oh okay then i can solve it i think
bahrom7893
  • bahrom7893
integrating factor is e^x, lemme try this will let u guys know how it goes
bahrom7893
  • bahrom7893
wait no its not, but at least now i know the expression, tryin to solve afk
anonymous
  • anonymous
here, this is called exact equation http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx
anonymous
  • anonymous
M(x,y) + N(x,y)dy/dx = 0
bahrom7893
  • bahrom7893
k
anonymous
  • anonymous
dont kill yourself bahrom this website is buggy, everytime i type something it freezes
anonymous
  • anonymous
still there?
anonymous
  • anonymous
have you done partial derivatives?
bahrom7893
  • bahrom7893
i have, but as i said my brains are not functionin at all, just too tired
anonymous
  • anonymous
we are looking for a function Fx (x,y) + Fy (x,y) dy/dx = 0, where Fx is the partial derivative of F(x,y) with respect to x (treat y as a constant) and Fy is the partial derivative of F(x,y) with respect to y (treat x as a constant)
anonymous
  • anonymous
ok .... :(
anonymous
  • anonymous
We have the equation M(x,y) dx + N(x,y) dy = 0 or M (x,y) + N(x,y)dy/dx = 0 M(x,y) = y and N(x,y) = 3 + 3x - y remember we can solve this and its called "exact" if M(x,y) = Fx (x,y) and N(x,y) = Fy (x,y), where Fx is the partial derivative of F(x,y) with respect to x (treat y as a constant) and Fy is the partial derivative of F(x,y) with respect to y (treat x as a constant) so if we have Fx (x,Y) + Fy (x,y) dy/dx = 0, then F(x,y) is our solution . ok so the last thing , assuming there exists such a F(x,y) it must be the case that Fxy = Fyx which is true for all continuous F(x,y). By substituting Fx = M and Fy = N we have My = Nx . So here Mx = 0 , and Ny = -1 . hmmm, not exact ok now we move to plan B and introduce an integrating factor http://www.cliffsnotes.com/study_guide/Integrating-Factors.topicArticleId-19736,articleId-19711.html
anonymous
  • anonymous
here is another website http://www.sosmath.com/diffeq/first/exact/exact.html and they have a non exact first order ode

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