## MathSofiya Group Title $y''+y'-2y=x+sin2x$ $y(0)=1$ $y'(0)=0$ auxiliary equation: $r^2+r-2=0$ $r_1=-2, r_2=1$ $y_c(x)=C_1e^{-2x}+C_2e^x$ $y''+y'-2y=x$ $y_{p1}(x)=Ax+B$ $y_{p1}'(x)=A$ $y_{p1}''(x)=0$ $A=-\frac12$ $B=-1$ $y_{p1}=-\frac12x-1$ one year ago one year ago

1. MathSofiya Group Title

Now on to yp_2, where I'm having some trouble...here is how far I've gotten.

2. MathSofiya Group Title

$y''+y'-2y=sin2x$ $y_{p2}(x)=Axcos2x+Bxsin2x$ $y_{p2}'(x)=2Acos2x-2Axsin2x+2Bsin2x+2Bxcos2x$ $y_{p2}''(x)=-4Asin2x-4Axcos2x+4Bcos2x-4Bxsin2x$

3. MathSofiya Group Title

@satellite73

4. satellite73 Group Title

really i am sorry i have no clue

5. psi9epsilon Group Title

there is no need to select Ax cos 2x + Bx sin 2x the reason is , if you take out x as a common factor, then that "x" is actually part of a solution so you are bound to get incorrect solutions

6. MathSofiya Group Title

So what should I do?

7. psi9epsilon Group Title

Did you try this using wronskian

8. MathSofiya Group Title

I don't know what that is

9. psi9epsilon Group Title

can you use wikipedia to understand that method, use the keywords "wronskian equation" or "wronskian determinant" I would use that approach to solve, I will use the method followed by you only under the cases where wronskian is 0

10. psi9epsilon Group Title

did you look up wronskian

11. across Group Title

You made a mistake assuming the form of $$y_{p_2}$$: I don't know if you've heard of the concept of an "annihilator" before, but the differential operator $$(D^2+4)$$ annihilates $$\sin2x$$. If we apply it to your ODE, we get:\begin{align}(D^2+4)(D^2+D-2)y&=(D^2+4)\sin2x,\m^2+4)(m^2+m-2)&=0,\\(m\pm2i)(m+2)(m-1)&=0.\end{align}Hence, \(y_{p_2}=A\cos2x+B\sin2x. This is different from what you got up there; notice that there is no $$x$$ multiplying neither $$\cos$$ nor $$\sin$$. Now try and try and wrap it up as you would from here.

12. MathSofiya Group Title

why is it $D^2+4$? Is it standard for the annihilator method or where did the 4 come from?

13. MathSofiya Group Title

I''m watching Khan's academy videos right now...I should have a better understanding in an hour or so

14. across Group Title

The differential operator$\left[D^2-2\alpha D+\left(\alpha^2+\beta^2\right)\right]^n$anihilates each of the functions$e^{\alpha x}\cos\beta x, xe^{\alpha x}\cos\beta x, \dots, x^{n-1}e^{\alpha x}\cos\beta x,\\e^{\alpha x}\sin\beta x, xe^{\alpha x}\sin\beta x, \dots, x^{n-1}e^{\alpha x}\sin\beta x.$

15. across Group Title

In your case, $$\alpha=0$$ and $$\beta=2$$.

16. TuringTest Group Title

I'm sure @across knows exactly what she's talking about and I don't, but I would just make a single $$y_p$$ which is the sum of the two individual ones

17. TuringTest Group Title

the guess for $$\cos(Ax)$$ or $$\sin(Ax)$$ is $$y_p=B\cos(Ax)+C\sin(Ax)$$

18. TuringTest Group Title

so add your two together and get$y_p=Ax+B+C\cos(2x)+D\cos(2x)$

19. TuringTest Group Title

not sure why you picked up the extra x on $$y _2$$

20. across Group Title

That's correct. It follows from the principle of superposition given differential operators are linear transformations. :) I was thrown off a little bit when I saw @MathSofiya formulating two particular solutions, but then told myself: "heck, it's the same thing anyway." So I went with her approach, but I'd also combine them. @TuringTest, that's right, but the term with the $$D$$ coefficient should be a $$\sin$$ (you could invert them, too, it doesn't matter).

21. TuringTest Group Title

@across oh yes, of of course, typo :P I'm also now remembering the annihilator thing you did, the need to add the 4 coming from differntiating sin(2x) twice and getting -4 right? (sorry to interject, just trying to remember)

22. MathSofiya Group Title

I still have two more videos to watch...brb, (hopefully it will make more sense then...)

23. TuringTest Group Title

or look quickly at example 8-d here (not that POMN is that thorough...)

24. TuringTest Group Title
25. TuringTest Group Title

as was said, it's the superposition principle at work is all

26. across Group Title

That's right, @TuringTest. Given $$\sin2x$$, applying the differential operator $$D^2+4$$ to it yields:\begin{align}(D^2+4)\sin2x&=D^2\sin2x+4\sin2x\\&=2D\cos2x+4\sin2x\\&=-4\sin2x+4\sin2x\\&=0.\end{align}:)

27. TuringTest Group Title

thanks! Gotta get sharp on that technique

28. MathSofiya Group Title

Let's start from $y''+y'-2y=x+sin2x$ and $y_p$ It's still not making much sense

29. TuringTest Group Title

if it were just g(x)=x our guess would be Yp=Ax+B, right?

30. MathSofiya Group Title

correct

31. TuringTest Group Title

and if it were just g(x)=sin(2x) what would the guess be?

32. MathSofiya Group Title

Asin2x+Bcos2x?

33. TuringTest Group Title

right, but let's call A=C and B=D so as not to confuse them with the other guess; they are not the same constants

34. MathSofiya Group Title

ok

35. TuringTest Group Title

so for g(x)=x+sin(2x) Yp=(Ax+B)+(Csin(2x)+Dsin(2x)) which is just the sum (superposition/linear combination) of our two guesses

36. TuringTest Group Title

since nothing on the right side is a linear combination f anything in the complimentary solution, your solutions are all linearly dependent i.e. we don't need to screw with them any more by adding more x's or anything

37. TuringTest Group Title

linearly independent*

38. TuringTest Group Title

hence the guess for the particular is$y_p=Ax+B+C\sin(2x)+B\cos(2x)$the guess for the g(x)=x part+ the guess for the g(x)=sin(2x) part making a little more sense (sorry that I seemed to have discouraged @across from giving her I'm sure much more insightful answer)

39. MathSofiya Group Title

I understand it now! Is that it?....or I still gotta solve for A B and C of course

40. across Group Title

No! That's totally fine. :) I couldn't have said it better.

41. TuringTest Group Title

thanks :) so yeah, now it's business as usual (certainly a bit tedious, but straightforward I think)

42. MathSofiya Group Title

A=1

43. TuringTest Group Title

I honestly don't feel like checking your work apart from wolfram, which is not always reliable, but hopefully you will agree.

44. MathSofiya Group Title

LOL....Yeah I got this from here on. Thanks Turing!

45. MathSofiya Group Title

Thanks to you too @across

46. MathSofiya Group Title

so I can skip the annihilator method (I rather avoid it because it looks confusing)

47. MathSofiya Group Title

?

48. TuringTest Group Title

it looks like wolfram wants to say A=-1/2 I will play with the system to find the constants, but I'm not going to rush it.

49. TuringTest Group Title

http://www.wolframalpha.com/input/?i=y''%2By'-2y%3Dx%2Bsin(2x)%2C%20y(0)%3D1%2C%20y'(0)%3D0&t=crmtb01 though I really don't trust wolf that much on DE's

50. MathSofiya Group Title

Hhmmmm....I'll do it manually too and compare

51. MathSofiya Group Title

I worked on understanding this idea for 2 days now...thanks guys for helping me finally understand it, honestly!

52. TuringTest Group Title

That's the best reward I could hope for, happy to help. I'll let you know what I get manually... eventually ;)

53. MathSofiya Group Title

No hurry =P

54. TuringTest Group Title

I did get A=-1/2 by hand

55. TuringTest Group Title

actually you can see that fairly quickly...

56. TuringTest Group Title

wanna see how?

57. MathSofiya Group Title

sure

58. MathSofiya Group Title

I forgot to take the first and second derivative before solving for A B and C...that's why I said A=1 OOOooops

59. TuringTest Group Title

you can also notice that the only part that will have an x left is Yp and since you have -2Yp you will have -2Ax on the left and x on the right hence A=-1/2 the rest is getting a bit frustrating :/

60. TuringTest Group Title

okay I have confirmed wolf's answer on paper

61. MathSofiya Group Title

I've gotten A=-1/2 B=-1/4 C=5/2 and D=-5/2

62. MathSofiya Group Title

I guess my C's and D's are wrong...oh well I'll take a closer look at wolfram

63. TuringTest Group Title

you wanna walk through it?

64. MathSofiya Group Title

Now I got A=-1/2 B=-1/4 C=-1/4 D=1/4 that's before considering y(0)=1 and y'(0)=0

65. TuringTest Group Title

nah, still got problems with C and D can you write out yp, yp', and yp'' ?

66. MathSofiya Group Title

$y_p=Ax+B+Csin2x+Dcos2x$ $y_p'=A+2Ccos2x-2Dsin2x$ $y_p''=-4Csin2x-4Dcos2x$

67. TuringTest Group Title

thanks now just write out the result of doing the sines and cosines in y''+y'-2y ignore the Ax an B as you have already solved it and it won't affect the C and D

68. MathSofiya Group Title

$(-4C-2D-2C)=1$ $(-4D+2C-2D)=0$

69. TuringTest Group Title

yep

70. MathSofiya Group Title

something is still wrong though

71. TuringTest Group Title

not yet, you just must have messed up solving the system

72. MathSofiya Group Title

oh I see what I did

73. TuringTest Group Title

sweet

74. MathSofiya Group Title

nope not yet C=3D and then you sub it into the first equation -6(3D)+2D=1 that still is D=-1/16

75. TuringTest Group Title

no it aint :P

76. TuringTest Group Title

you changed a sign

77. TuringTest Group Title

the first equation is -2D, not +2D

78. MathSofiya Group Title

that's what I had right? $(-4C-2D-2C)=1$ $(-4D+2C-2D)=0$ simplified $(-6C-2D)=1$ $(-6D+2C)=0$ multiply the lower equation by 3 $(-6C-2D)=1$ $(+6C-18D)=0$ -----------------------+ -20D=1 D=-1/20

79. TuringTest Group Title

yay!

80. MathSofiya Group Title

LATEX makes all the difference!

81. TuringTest Group Title

I really think that if programs are meant to do anything tedious in math, systems are about the most perfect example there is although, a few linear algebra tricks can solve these 2x2's pretty darn fast

82. TuringTest Group Title

btw I made the $$exact$$ same mistake and got D=1/16 first

83. TuringTest Group Title

the system to find the constants in the particular is actually a bit worse I'd say, so I used some linear algebra stuff there

84. TuringTest Group Title

I mean in the complimentary

85. across Group Title

Hopefully this summarizes everything that went on in here nicely for you :) https://dl.dropbox.com/u/12189012/OS.pdf

86. MathSofiya Group Title

oh my gosh! did you just make that?

87. across Group Title

I felt like practicing my rusty $$\LaTeX$$ skills.

88. TuringTest Group Title

That is quite generous, and looks much nicer than my two pages of scribbles with things crossed out ;)

89. MathSofiya Group Title

you are soo sweet! thank you!

90. MathSofiya Group Title

OS rocks!

91. TuringTest Group Title

yes it does :)