dfresenius Group Title xy' + y = −2xy^2 Bernoulli Differential equation. one year ago one year ago

1. dfresenius Group Title

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

2. dfresenius Group Title

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

3. dfresenius Group Title

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4. dfresenius Group Title

i suppose i should state what bernoulli is

5. zepdrix Group Title

$\large xy' + y = −2xy^2$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 Group Title

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

7. dfresenius Group Title

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8. abb0t Group Title

Haven't seen bernoulli in some time.

9. dfresenius Group Title

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 Group Title

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

11. abb0t Group Title

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

12. abb0t Group Title

after you substitute everything.

13. zepdrix Group Title

$\large y'+p(x)y=q(x)y^n$ 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 Group Title

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

15. dfresenius Group Title

ya, im just puzzled

16. dfresenius Group Title

so, with my equation,

17. zepdrix Group Title

$\large xy' + y = −2xy^2$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. dfresenius Group Title

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19. zepdrix Group Title

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. dfresenius Group Title

ok that makes sense

21. dfresenius Group Title

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22. zepdrix Group Title

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. dfresenius Group Title

oh the derivative gotcha

24. zepdrix Group Title

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. dfresenius Group Title

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26. zepdrix Group Title

Yes good! :)

27. zepdrix Group Title

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

28. dfresenius Group Title

ok

29. dfresenius Group Title

should i just switch sides?

30. zepdrix Group Title

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 Group Title

Just multiply everything by -1 :)

32. dfresenius Group Title

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33. zepdrix Group Title

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

34. zepdrix Group Title

35. dfresenius Group Title

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36. zepdrix Group Title

heh XD

37. dfresenius Group Title

ok so, i integrate both sides?

38. zepdrix Group Title

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

39. dfresenius Group Title

ya let me check real quick

40. dfresenius Group Title

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41. zepdrix Group Title

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

42. dfresenius Group Title

dangit, i need to stop rushing

43. zepdrix Group Title

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 Group Title

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

45. dfresenius Group Title

haha nice

46. dfresenius Group Title

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47. zepdrix Group Title

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

48. dfresenius Group Title

1/x

49. zepdrix Group Title

Cool c:

50. dfresenius Group Title

51. dfresenius Group Title

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

52. zepdrix Group Title

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. dfresenius Group Title

ok,

54. dfresenius Group Title

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55. dfresenius Group Title

now, i want to isolate u' right?

56. zepdrix Group Title

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 Group Title

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

58. zepdrix Group Title

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

59. dfresenius Group Title

yes, so I can simplify the left

60. zepdrix Group Title

ya sounds good c:

61. dfresenius Group Title

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62. zepdrix Group Title

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

63. zepdrix Group Title

yah everything looks good so far.

64. dfresenius Group Title

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65. dfresenius Group Title

Cx

66. zepdrix Group Title

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

67. dfresenius Group Title

ok, u=y^-1

68. zepdrix Group Title

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

69. dfresenius Group Title

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70. zepdrix Group Title

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. dfresenius Group Title

YESSS

72. dfresenius Group Title

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

73. zepdrix Group Title

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