anonymous
  • anonymous
Differential Equations(Integration Factor): show that (x+y)^(-2) is the integration factor of [x^(2)+2xy-y^(2)]dx + [y^(2)+2xy-x^(2)]dy = 0 Need Help, thank you very much.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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danica518
  • danica518
http://38.media.tumblr.com/1741e6ccb28233179cc63040cdf09524/tumblr_n2hhznE3Ub1rfduvxo1_250.gif
danica518
  • danica518
is this the given equation exact
anonymous
  • anonymous
I think it is non-exact since the integration factor is given :)

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danica518
  • danica518
take df/dx = x^(2)+2xy-y^(2) df/dy=y^(2)+2xy-x^(2)
anonymous
  • anonymous
getting the partial derivatives results to M'(y)=2x-2y;N'(x)=2y-2x
danica518
  • danica518
no
anonymous
  • anonymous
which can be summarized to M'(y)=-N'(x), but i want to know how did (x+y)^(-2) is solved
danica518
  • danica518
yes
danica518
  • danica518
:)
anonymous
  • anonymous
oh.... so, how did the integration factor became (x+y)^(2)? that is what i can't understand ^^;
anonymous
  • anonymous
i mean (x+y)^(-2)
freckles
  • freckles
well have you tried dividing both sides by (x+y)^2 and checking if M_y=N_x?
freckles
  • freckles
that is all you need to show that it is the integrating factor you need
danica518
  • danica518
you gotta just show the integration is true here, not how its found,
danica518
  • danica518
im curious how ud go about finding that integration factor here too
danica518
  • danica518
it might from like a sturmlouiville form t
freckles
  • freckles
\[\text{ if } \frac{1}{N}[M_y-N_x]=h(x) \text{ then the integrating factor is } e^{\int\limits h(x) dx} \\ \text{ if } \frac{1}{M}[N_x-M_y]=k(y) \text{ then the integrating factor is } e^{\int\limits k(y) dy}\]
freckles
  • freckles
where you have the differential equation:\[Mdx+Ndy=0\]
danica518
  • danica518
okay
anonymous
  • anonymous
@freckles \[M=1-2[\frac{ \frac{ y }{ x } }{ 1+\frac{ y }{ x } }]^{2} ; N=1-2[\frac{1}{ 1+\frac{ y }{ x } }]^{2}\]
anonymous
  • anonymous
that's what i got when i divide them with (x+y)^(2) ^^;
danica518
  • danica518
show the newequation u get after multiplying by integration factor is still exact
freckles
  • freckles
\[M=x^2+2xy-y^2 \\ M=x^2+2xy+y^2-2y^2 \\ M=(x+y)^2-2y^2 \\ \frac{M}{(x+y)^2}=1-\frac{2y^2}{(x+y)^2} \\ N=y^2+2xy-x^2 \\ N=y^2+2xy+x^2-2x^2 \\ N=(x+y)^2-2x^2 \\ \frac{N}{(x+y)^2}=1-\frac{2x^2}{(x+y)^2}\]
freckles
  • freckles
\[\text{ new } M=1-\frac{2y^2}{(x+y)^2} \\ \text{ new} N=1-\frac{2x^2}{(x+y)^2}\]
freckles
  • freckles
by the way those two things I mentioned won't work since you don't get a function of x or a function of y you get a function of x and y for either try but anyways all you need to do is show new M_y=new N_x
anonymous
  • anonymous
the point of an integration factor is to turn an 'inexact' differential \(P\, dx+M\, dy\) into an 'exact' or total one \(d\Phi=\mu P\, dx+\mu M\, dy\), which is equivalent to solving the following PDE: $$\frac{\partial}{\partial y}[\mu P]=\frac{\partial}{\partial x}[\mu M]$$
anonymous
  • anonymous
since we want $$\mu P=\frac{\partial\Phi}{\partial x},\mu M=\frac{\partial\Phi}{\partial y}$$ and \(\dfrac{\partial^2\Phi}{\partial y\,\partial x}=\dfrac{\partial^2\Phi}{\partial x\,\partial y}\)
anonymous
  • anonymous
anyways, the standard solution presumes \(\mu\) is a function of a single variable, either \(\mu(x)\) or \(\mu(y)\) so $$\frac{\partial}{\partial y}[\mu P]=\frac{\partial\mu}{\partial y}P+\mu\frac{\partial P}{\partial y}\\\frac{\partial}{\partial x}[\mu M]=\frac{\partial\mu}{\partial x}M+\mu\frac{\partial M}{\partial x}$$ so $$\frac{\partial\mu}{\partial y}P+\mu\frac{\partial P}{\partial y}=\frac{\partial\mu}{\partial x}M+\mu\frac{\partial M}{\partial x}\\P\frac{\partial\mu}{\partial y}-M\frac{\partial\mu}{\partial x}+\left(\frac{\partial P}{\partial y}-\frac{\partial M}{\partial x}\right)\mu=0$$ which then reduces to a linear first-order ODE if we choose either \(\mu_y=\frac{d\mu}{dy},\mu_x=0\) or \(\mu_x=\frac{d\mu}{dx},\mu_y=0\)
anonymous
  • anonymous
Thanks for the help everyone :) i haven't finished solving yet but i got the concept. Thanks again.

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