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.

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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
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http://38.media.tumblr.com/1741e6ccb28233179cc63040cdf09524/tumblr_n2hhznE3Ub1rfduvxo1_250.gif
is this the given equation exact
I think it is non-exact since the integration factor is given :)

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take df/dx = x^(2)+2xy-y^(2) df/dy=y^(2)+2xy-x^(2)
getting the partial derivatives results to M'(y)=2x-2y;N'(x)=2y-2x
no
which can be summarized to M'(y)=-N'(x), but i want to know how did (x+y)^(-2) is solved
yes
:)
oh.... so, how did the integration factor became (x+y)^(2)? that is what i can't understand ^^;
i mean (x+y)^(-2)
well have you tried dividing both sides by (x+y)^2 and checking if M_y=N_x?
that is all you need to show that it is the integrating factor you need
you gotta just show the integration is true here, not how its found,
im curious how ud go about finding that integration factor here too
it might from like a sturmlouiville form t
\[\text{ if } \frac{1}{N}[M_y-N_x]=h(x) \text{ then the integrating factor is } e^{\int\limits h(x) dx} \\ \text{ if } \frac{1}{M}[N_x-M_y]=k(y) \text{ then the integrating factor is } e^{\int\limits k(y) dy}\]
where you have the differential equation:\[Mdx+Ndy=0\]
okay
@freckles \[M=1-2[\frac{ \frac{ y }{ x } }{ 1+\frac{ y }{ x } }]^{2} ; N=1-2[\frac{1}{ 1+\frac{ y }{ x } }]^{2}\]
that's what i got when i divide them with (x+y)^(2) ^^;
show the newequation u get after multiplying by integration factor is still exact
\[M=x^2+2xy-y^2 \\ M=x^2+2xy+y^2-2y^2 \\ M=(x+y)^2-2y^2 \\ \frac{M}{(x+y)^2}=1-\frac{2y^2}{(x+y)^2} \\ N=y^2+2xy-x^2 \\ N=y^2+2xy+x^2-2x^2 \\ N=(x+y)^2-2x^2 \\ \frac{N}{(x+y)^2}=1-\frac{2x^2}{(x+y)^2}\]
\[\text{ new } M=1-\frac{2y^2}{(x+y)^2} \\ \text{ new} N=1-\frac{2x^2}{(x+y)^2}\]
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
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: $$\frac{\partial}{\partial y}[\mu P]=\frac{\partial}{\partial x}[\mu M]$$
since we want $$\mu P=\frac{\partial\Phi}{\partial x},\mu M=\frac{\partial\Phi}{\partial y}$$ and \(\dfrac{\partial^2\Phi}{\partial y\,\partial x}=\dfrac{\partial^2\Phi}{\partial x\,\partial y}\)
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 \(\mu_x=\frac{d\mu}{dx},\mu_y=0\)
Thanks for the help everyone :) i haven't finished solving yet but i got the concept. Thanks again.

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