## anonymous one year ago 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.

1. danica518
2. danica518

is this the given equation exact

3. anonymous

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

4. danica518

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

5. anonymous

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

6. danica518

no

7. anonymous

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

8. danica518

yes

9. danica518

:)

10. anonymous

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

11. anonymous

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

12. freckles

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

13. freckles

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

14. danica518

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

15. danica518

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

16. danica518

it might from like a sturmlouiville form t

17. freckles

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

18. freckles

where you have the differential equation:$Mdx+Ndy=0$

19. danica518

okay

20. anonymous

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

21. anonymous

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

22. danica518

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

23. freckles

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

24. freckles

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

25. 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

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

27. anonymous

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

28. 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 $$\mu_x=\frac{d\mu}{dx},\mu_y=0$$

29. anonymous

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