## anonymous 3 years ago $(2y^2 - 6xy)dx + (3xy-4x^2)dy =0$ Find an integrating factor of the form $$x^ny^m$$ and solve the equation...

1. abb0t

1st check to make sure that they are exact.

2. anonymous

$$M = 2y^2 - 6xy$$ $$N= 3xy-4x^2$$ $$\frac{\partial M}{\partial y}=4y - 6x$$ $$\frac{\partial N}{\partial x}=3y-8x$$ $\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=\frac{3y-8x-(4y - 6x)}{2y(y-3x)}$Doesn't go right..

3. abb0t

$\frac{ ∂M }{ ∂x }=\frac{ ∂N }{ ∂y }$

4. abb0t

check your partial derivatives. they should match up

5. anonymous

either that or your teacher is horrible and wants you to solve this another way ll

6. UnkleRhaukus

integrating factor!

7. anonymous

But I couldn't find a suitable integrating factor... :'(

8. anonymous

ah yes this is not an exact solution one

9. abb0t

UnkleR is right. Find the integrating factor to make it exact.

10. anonymous

That's why I tried that partial.. - partial.. / ... to find one..

11. anonymous

is that a hint... try getting one in such that it looks like above

12. abb0t

You were on the right path.

13. abb0t

$\frac{ M_y-N_x }{ -M }$

14. abb0t

15. anonymous

The way I learn to get an integrating factor is by showing $$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}$$ or $$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}$$ equals to a function of x/y .. But for this (and the next question), I got troubles..

16. UnkleRhaukus

$\frac{\partial M}{\partial y}\neq \frac{\partial N}{\partial x}$ $R=R(x,y)=x^ny^m$ $\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}$ $R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}$

17. anonymous

How... does... that... work...?

18. UnkleRhaukus

which bit

19. anonymous

$\frac{\partial R(x,y)M}{\partial y}= \frac{\partial R(x,y)N}{\partial x}$And $R_y(x,y)M+R(x,y)\frac{\partial M}{\partial y}=R_x(x,y)N+R(x,y)\frac{\partial N}{\partial x}$

20. UnkleRhaukus

well we want a integrating factor that will make the partial derivatives equal (which will makes the equation exact) applying the product rule for derivatives on both sides we get that last line i have used mixed notation for derivatives for some reason $Z_w\leftrightarrow\frac{\partial Z}{\partial w}$(these mean the same thing, just written differently )

21. UnkleRhaukus

so you know $R,M,N$substitute these in

22. UnkleRhaukus

take the partial derivatives

23. anonymous

Is R the integrating factor?

24. UnkleRhaukus

yes

25. anonymous

So, I should multiply the equation by R, right?

26. anonymous

$R=x^ny^m$$M=2y^2-6xy$$N=3xy-4x^2$ $\frac{\partial R}{\partial y} M + \frac{\partial M}{\partial y} R= \frac{\partial R}{\partial x} N+ \frac{\partial N}{\partial x}R$ $LS$$=mx^ny^{m-1}(2y^2-6xy) + (4y-6x)x^ny^m$$=2mx^ny^{m+1}-6mx^{n+1}y^m + 4y^{m+1}x^n – 6x^{n+1}y^m$$(2m+4)x^ny^{m+1}-(6m+1)x^{n+1}y^m$ $RS$$=nx^{n-1}y^m(3xy-4x^2) + (3y-8x)x^ny^m$$=3nx^ny^{m+1}-4nx^{n+1}y^m+3x^ny^{m+1} – 8x^{n+1}y^m$$(3n+3)x^ny^{m+1} – (4n+8)x^{n+1}y^m$ So, $(2m+4)x^ny^{m+1}-(6m+1)x^{n+1}y^m = (3n+3)x^ny^{m+1} – (4n+8)x^{n+1}y^m$ 2m+4 = 3n+3 6m+1 = 4n+8 m=5/2 , n=2 But that is not right :\

27. UnkleRhaukus

hmm, i cannot see any error in your working , how do you know its not right?

28. anonymous

Because the book whispered me the answer :(

29. UnkleRhaukus

$\tiny \text{can you whisper it to me too please , }$

30. anonymous

$\tiny \color{white}{\mu =xy}$

31. UnkleRhaukus

i see

32. hartnn

$$\huge \color{red}{\mu =xy}$$ now, i see also.

33. UnkleRhaukus

well we were close $x^2y^{5/2}(2y^2 - 6xy)\text dx + x^2y^{5/2}(3xy-4x^2)\text dy =0$ $(2x^2y^{9/2} - 6x^3y^{7/2})\text dx +(3x^3y^{7/2}-4x^4y^{5/2})\text dy =0$ $\frac{\partial MR}{\partial y}=9x^2y^{7/2}-21x^3y^{5/2}$ $\frac{\partial NR}{\partial x}=9x^2y^{7/2}-16x^3y^{5/2}$

34. UnkleRhaukus

$21\sim16$

35. anonymous

I.... don't .... understand.... :(

36. UnkleRhaukus

$R=\mu$

37. anonymous

Hmm... I should do it all over again?!

38. anonymous

xy vs x^2 y^(5/2) They look too different to me!

39. hartnn

life would be so much easier if we could use substitution...and not IF.

40. anonymous

Then, this, again, proves that life is not easy :(

41. UnkleRhaukus

well i used 6 bits of paper , and got the same answer we had before, bother

42. anonymous

6... bits?! of paper?! :(

43. hartnn

$$LS=(2m+4)x^ny^{m+1}-(6m+6)x^{n+1}y^m$$

44. hartnn

checking RS

45. hartnn

2m+4 = 3n+3 6m+6 = 4n+8 m=1 n=1

46. hartnn

just one silly mistale!

47. hartnn

*mistake

48. UnkleRhaukus

oh

49. UnkleRhaukus

i see it now , such a tiny tiny mistale

50. hartnn

thank you so much @UnkleRhaukus , i learned a new method today...R=x^m y^n

51. UnkleRhaukus

OK, so now we have the integrating factor $R(x,y)=\mu(x,y)=xy$, lets integrate!

52. hartnn

actually , i wanted to see the look at @RolyPoly face, when he/she finds out how $$\tiny tiny$$ the error was :P

53. UnkleRhaukus

$\int \mu N\text dx$$=\int (2xy^3−6x^2y^2)\text dx$$=x^2y^3-2x^3y^2+g(y)$ $\int\mu M\text dy$$=\int(3x^2y^2−4x^3y)\text dy$$=x^2y^3-2x^3y^2+h(x)$ $\implies g(y)=h(x)=0$ $f(x,y)=x^2y^3-2x^3y^2=c$

54. UnkleRhaukus

or the other way $f(x,y)=\int \mu N\text dx=x^2y^3-2x^3y^2+g(y)$ $\frac{f(x,y)}{\text dy}=3x^2y^2-4x^3y+g'(y)=\mu M=3x^2y^2−4x^3y$ $\implies g'(y)=0$$g(y)=c_1$ $f(x,y)=x^2y^3-2x^3y^2+c_1=0$

55. UnkleRhaukus

$c_1=-c$

56. UnkleRhaukus

woo!

57. UnkleRhaukus

wasn't that fun?

58. hartnn

$\begin{array}l\color{red}{\text{y}}\color{orange}{\text{e}}\color{#e6e600}{\text{s}}\color{green}{\text{,}}\color{blue}{\text{ }}\color{purple}{\text{o}}\color{purple}{\text{f}}\color{red}{\text{c}}\color{orange}{\text{o}}\color{#e6e600}{\text{u}}\color{green}{\text{r}}\color{blue}{\text{s}}\color{purple}{\text{e}}\color{purple}{\text{ }}\color{red}{\text{:}}\color{orange}{\text{)}}\color{#e6e600}{\text{}}\end{array}$

59. anonymous

Mummy!!!! I want to die :'(

60. hartnn

lol :P

61. hartnn

plz don't die.

62. anonymous

She! It's a she!

63. anonymous

Thanks for rescue!!!! I'm sorry for my silly mistake!!

64. hartnn

$\begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#e6e600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array}$

65. UnkleRhaukus

*mistale

66. anonymous

Hmmm.. So must I use $\frac{\partial R}{\partial y} M + \frac{\partial M}{\partial y} R= \frac{\partial R}{\partial x} N+ \frac{\partial N}{\partial x}R$ to find R?! You know it's a pain :(

67. hartnn

yes, it is pain, but there is different kind of pleasure when we arrive at correct answer after all that!

68. anonymous

I must take this pain then.. Thanks again for all of your help!! Much appreciated!! (Btw, who wants the medal?)

69. UnkleRhaukus

if you were wondering what the solution 'looks' like ,...

70. UnkleRhaukus

solutions*

71. anonymous

Ugly :( How do you get the plot?

72. UnkleRhaukus

i used a graphing program ,

73. anonymous

That is..?