Question

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.

Answer 1

http://38.media.tumblr.com/1741e6ccb28233179cc63040cdf09524/tumblr_n2hhznE3Ub1rfduvxo1_250.gif

Answer 2

is this the given equation exact

Answer 3

I think it is non-exact since the integration factor is given :)

Answer 4

take df/dx = x^(2)+2xy-y^(2)
df/dy=y^(2)+2xy-x^(2)

Answer 5

getting the partial derivatives results to M'(y)=2x-2y;N'(x)=2y-2x

Answer 6

no

Answer 7

which can be summarized to M'(y)=-N'(x), but i want to know how did (x+y)^(-2) is solved

Answer 8

yes

Answer 9

:)

Answer 10

oh.... so, how did the integration factor became (x+y)^(2)? that is what i can't understand ^^;

Answer 11

i mean (x+y)^(-2)

Answer 12

well have you tried dividing both sides by (x+y)^2
and checking if M_y=N_x?

Answer 13

that is all you need to show that it is the integrating factor you need

Answer 14

you gotta just show the integration is true here, not how its found,

Answer 15

im curious how ud go about finding that integration factor here too

Answer 16

it might from like a sturmlouiville form t

Answer 17

\[\text{ if } \frac{1}{N}[M_y-N_x]=h(x) \text{ then the integrating factor is } e^{\int\limits h(x) dx} \\ \text{ if } \frac{1}{M}[N_x-M_y]=k(y) \text{ then the integrating factor is } e^{\int\limits k(y) dy}\]

Answer 18

where you have
the differential equation:\[Mdx+Ndy=0\]

Answer 19

okay

Answer 20

@freckles \[M=1-2[\frac{ \frac{ y }{ x } }{ 1+\frac{ y }{ x } }]^{2} ; N=1-2[\frac{1}{ 1+\frac{ y }{ x } }]^{2}\]

Answer 21

that's what i got when i divide them with (x+y)^(2) ^^;

Answer 22

show the newequation u get after multiplying by integration factor is still exact

Answer 23

\[M=x^2+2xy-y^2 \\ M=x^2+2xy+y^2-2y^2 \\ M=(x+y)^2-2y^2 \\ \frac{M}{(x+y)^2}=1-\frac{2y^2}{(x+y)^2} \\ N=y^2+2xy-x^2 \\ N=y^2+2xy+x^2-2x^2 \\ N=(x+y)^2-2x^2 \\ \frac{N}{(x+y)^2}=1-\frac{2x^2}{(x+y)^2}\]

Answer 24

\[\text{ new } M=1-\frac{2y^2}{(x+y)^2} \\ \text{ new} N=1-\frac{2x^2}{(x+y)^2}\]

Answer 25

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

Answer 26

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: $$\frac{\partial}{\partial y}[\mu P]=\frac{\partial}{\partial x}[\mu M]$$

Answer 27

since we want $$\mu P=\frac{\partial\Phi}{\partial x},\mu M=\frac{\partial\Phi}{\partial y}$$ and \(\dfrac{\partial^2\Phi}{\partial y\,\partial x}=\dfrac{\partial^2\Phi}{\partial x\,\partial y}\)

Answer 28

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 \(\mu_x=\frac{d\mu}{dx},\mu_y=0\)

Answer 29

Thanks for the help everyone :) i haven't finished solving yet but i got the concept. Thanks again.