## anonymous 3 years ago xy' + y = −2xy^2 Bernoulli Differential equation.

1. anonymous

Yes, this is a bit strange because they want me to substitute u=y^(1-n)

2. anonymous

so, im thinking to start off, i will get x and y on one side.

3. anonymous

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4. anonymous

i suppose i should state what bernoulli is

5. zepdrix

$\large xy' + y = −2xy^2$Hmm it's been a lil while.. I'm trying to remember how to do these. I think you want to get the equation to this form. ~Divide both sides by x, Divide both sides by y^2. $\large y^{-2}y' + \frac{1}{x}y^{-1} = −2$ Then let $$\large u=y^{-1}$$. Hmm $$\large u'=-y^{-2}$$

6. zepdrix

Woops, $$\large u'=-y^{-2}y'$$

7. anonymous

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8. abb0t

Haven't seen bernoulli in some time.

9. anonymous

ya im a bit shaky on this but im caught up to you zep, so the u'=y'y^-2 , can you explain that?

10. abb0t

u' = y^-2y' is what your going to substitute. essentially, you're going to get a linear differential euqation.

11. abb0t

You should get something like: u' + p(x)u=g(x)

12. abb0t

after you substitute everything.

13. zepdrix

$\large y'+p(x)y=q(x)y^n$ Dividing both sides by y^n, $$\large \color{royalblue}{y^{-n}y'}+p(x)\color{orangered}{y^{1-n}}=q(x)$$ Making the sub $$\large \color{orangered}{u=y^{1-n}}$$ $$\large u'=(1-n)y^{-n}y' \qquad \rightarrow \qquad \color{royalblue}{\dfrac{1}{1-n}u'=y^{-n}y'}$$ Giving us, yah something like what abbot said :)

14. zepdrix

I tried to use colors to make that clearer, but it's still rather confusing lol

15. anonymous

ya, im just puzzled

16. anonymous

so, with my equation,

17. zepdrix

$\large xy' + y = −2xy^2$Dividing each side by x gives us,$\large y' + \frac{1}{x}y = −2y^2$Dividing each side by y^2 (Because that is our y^n is this case) gives,$\large y^{-2}y' + \frac{1}{x}\color{royalblue}{y^{1-2}} = −2$Then we want to make that blue term our $$\large u$$.

18. anonymous

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19. zepdrix

Our $$\large n$$ in this case is 2, because our y^n was the term on the RIGHT side of the equation. $$\large u=y^{1-n} \qquad \rightarrow \qquad u=y^{2-1}\qquad \rightarrow \qquad u=y^{-1}$$

20. anonymous

ok that makes sense

21. anonymous

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22. zepdrix

I think maybe you missed a negative when taking the derivative of your u. $$\huge u=y^{-1} \qquad \rightarrow \qquad u'=\color{red}{-}y^{-2}y'$$

23. anonymous

oh the derivative gotcha

24. zepdrix

I assume we're taking the derivative with respect to $$\large x$$. That's probably why we get a prime term on each side :d

25. anonymous

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26. zepdrix

Yes good! :)

27. zepdrix

Remember that before you can get an integrating factor and all that fun stuff, your leading term needs to be positive.

28. anonymous

ok

29. anonymous

should i just switch sides?

30. zepdrix

Or maybe a better way to say that would be, it needs to be in standard form* no coefficients or anything on the leading term.

31. zepdrix

Just multiply everything by -1 :)

32. anonymous

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33. zepdrix

Woops, standard form looks like this,$\large u'+P(x)u=Q(x)$

34. zepdrix

35. anonymous

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36. zepdrix

heh XD

37. anonymous

ok so, i integrate both sides?

38. zepdrix

No you need to find an integrating factor $$\large \mu=e^{\int P(x)\;dx}$$ Look familiar I hope? :)

39. anonymous

ya let me check real quick

40. anonymous

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41. zepdrix

Woops the negative sign is also part of your $$\large P(x)$$. Recall that the standard form shows addition in the middle there.

42. anonymous

dangit, i need to stop rushing

43. zepdrix

And your equation should be $$\large u'-\dfrac{1}{x}u=2$$ Positive 2 on the right. I think we missed that earlier.

44. zepdrix

Yes you do need to stop rushing Chris Farley!! Slow down in life! Or you'll end up leaving us too soon......... :'C

45. anonymous

haha nice

46. anonymous

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47. zepdrix

$\huge e^{-\ln x} \qquad = \qquad e^{\ln (x^{-1})} \qquad = \qquad ?$

48. anonymous

1/x

49. zepdrix

Cool c:

50. anonymous

51. anonymous

ok so now that i have that, i multiply by x^-1

52. zepdrix

Yes, if you want to be thorough, then multiply through by your integrating factor. I usually skip that step. But it might be good not to skip it since you're learning it still.

53. anonymous

ok,

54. anonymous

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55. anonymous

now, i want to isolate u' right?

56. zepdrix

Looks good so far c: BTW, just a little note: You can scale the size of the drawing box if you don't want to take up that much space.|dw:1362711683197:dw|

57. zepdrix

From here, you want to recognize that the LEFT side is the result of the product rule.

58. zepdrix

$\large \left[u(x)f(x)\right]' \qquad = \qquad u'(x)f(x)+u(x)f'(x)$ We do that have in this case don't we? Where our f(x)=x^{-1}

59. anonymous

yes, so I can simplify the left

60. zepdrix

ya sounds good c:

61. anonymous

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62. zepdrix

Oh sorry I forget you were calling it $$\large I$$. I hope I didn't make that more confusing c:

63. zepdrix

yah everything looks good so far.

64. anonymous

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65. anonymous

Cx

66. zepdrix

Yes good. Then we have to do a little work changing our $$\large u$$ back to $$\large y$$

67. anonymous

ok, u=y^-1

68. zepdrix

Good.$\large \frac{1}{y}=2x \ln x+Cx$So how do we solve for y?

69. anonymous

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70. zepdrix

Yay good job! I threw it into Wolfram just a moment ago to make sure we didn't make any fatal errors. http://www.wolframalpha.com/input/?i=xy%27%2By%3D-2xy%5E2 Same result! Yay! :)

71. anonymous

YESSS

72. anonymous

thank you so much for your patience, i have been doing very well in my class because of your help

73. zepdrix

Oh that's good to hear! c: I'm sure it's not just me. You seem to understand these concepts very very quickly.