## anonymous 4 years ago Solve using variation of parameters $y''-2y'+y=e^{2x}$

1. TuringTest

complimentary....

2. anonymous

meaning?

3. anonymous

$y_c=c_1e^{-x}+c_2xe^{-x}$

4. TuringTest

right...

5. TuringTest

now we need the wronskian, do you know about that?

6. anonymous

nope

7. TuringTest

it is the determinant of a matrix

8. TuringTest

for$y_c=c_1y_1+c_2y_2$the Wronskian is$W(y_1,y_2)=\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|$

9. TuringTest

that's a determinant in case you forgot, so find that determinant

10. anonymous

$\left|\begin{matrix}e^{-x}&xe^{-x}\\-e^{-x}&-e^{-x}(x-1)\end{matrix}\right|$ =$e^{-x}(e^{-x}(x-1))-xe^{-x}e^{-x}$

11. anonymous

completely wrong?

12. anonymous

oh it's plus

13. TuringTest

yep you caught it :)

14. anonymous

$e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}$

15. anonymous

that's it?

16. TuringTest

you dropped a minus

17. anonymous

where?

18. anonymous

ohhhh

19. anonymous

$-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}$

20. TuringTest

right

21. TuringTest

simplify and be happy :)

22. anonymous

does it equal$e^{-x}$? $-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}$ ${-x}(e^{-x}(x-1))+xe^{-x}=1$

23. anonymous

$(e^{-x}(x-1))+xe^{-x}=1$

24. anonymous

typo

25. TuringTest

no, why did you set it equal to e^-x, we can't set it equal to anything yet...

26. anonymous

27. anonymous

so how should I simplify it then?

28. TuringTest

$-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}$now try again while I try to eat ;)

29. anonymous

-(x-1)=1-x 1-x=1-x

30. anonymous

no no no ...wait a second

31. anonymous

now i get 2-x=2-x

32. anonymous

nope still wrong

33. anonymous

x=1

34. anonymous

stop laughing

35. anonymous

Let's try that again $-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}$

36. anonymous

they're just equal to each other

37. anonymous

$-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}$

38. anonymous

$-e^{-2x}(x-1)+xe^{-2x}$

39. anonymous

$e^{-2x}((1-x)+x)$

40. anonymous

$e^{-2x}$

41. anonymous

yes?

42. TuringTest

yes, sorry, I was afk

43. anonymous

LOL ok

44. TuringTest

do you know the formula for the particular solution for variation of parameters?

45. anonymous

46. anonymous

something like that?

47. TuringTest

almost unidentifiable to me in that form, but correct I'm sure I have the more direct formula memorized http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx (the green box with the formula that says "variation of parameters)

48. anonymous

is the first one a y''? $y''+q(t)y'+r(t)y=g(t)$

49. TuringTest

yeah, it looks screwy I know

50. anonymous

do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like

51. TuringTest

yes and in this case they are the coefficients, which in this case are constants the formula is just pointing out that they may not be

52. anonymous

oh I see, soo... $y''+q(t)y'+r(t)y=g(t)$ $y''-2y'+y=e^{2x}$ so now I integrate

53. TuringTest

right, using the formula... r(t) and r(t) don't even come into the formula for the particular solution in variation of parameters

54. TuringTest

q(t) and r(t)... they determine the complimentary only, which as you can see *is* part of the formula for the particular

55. anonymous

$Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{-x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx$

56. TuringTest

the first integrand is wrong I think

57. TuringTest

I think you put the wrong g(x)

58. anonymous

$Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx$

59. TuringTest

oh wait, wait, the denominator in each integrand is the wronskian, which we found to be e^(-2x) I think you are confusing notation with the parentheses

60. anonymous

Oh lol so the denominator is just e^{-2x}

61. TuringTest

right

62. TuringTest

notice that makes our integrals quite tolerable :)

63. anonymous

$Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}}dx$ That's a lot easier to integrate

64. TuringTest

oh yeah :) especially after a little simplification

65. anonymous

$-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}$

66. anonymous

so what was the point of doing variation of parameters

67. TuringTest

well now you have a particular solution

68. TuringTest

y(x)=yc+yp you got you now so you're good to go

69. TuringTest

you got yp*

70. TuringTest

I did not check your integral btw

71. anonymous

that's ok I did wolfram

72. TuringTest

right-o, I'd do the same right now :)

73. anonymous

oh I see $c_1e^{-x}+c_2xe^{-x}-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}$

74. TuringTest

yep, and of course if this is an IVP we now proceed to find c1 and c2 otherwise we're done having found the general solution

75. anonymous

good, wow, that was a lot THanks Turing!

76. TuringTest

Always a pleasure Sofiya :D

77. anonymous

Looking back over this, I don't understand how you guys got a negative double root to the characteristic equation. I got m_1 = m_2 = +1. Plugging into the normal formula for repeated roots, getting e^x and xe^x, not e^-x. Am I missing something? @ganeshie8

78. anonymous

@jim_thompson5910 @agent0smith

79. anonymous

@Zarkon @terenzreignz Is anybody else familiar with ODE's seeing this, or am I missing something?

80. anonymous

@zepdrix Anybody?

81. zepdrix

@Mendicant_Bias For the homogeneous solution? Yes the characteristic equation appears to be giving +1 as a repeated root, as you indicated

82. zepdrix

So yah, that was a mistake early in the thread :C