## MathSofiya 3 years ago $y''+y'-2y=x+sin2x$ $y(0)=1$ $y'(0)=0$ auxiliary equation: $r^2+r-2=0$ $r_1=-2, r_2=1$ $y_c(x)=C_1e^{-2x}+C_2e^x$ $y''+y'-2y=x$ $y_{p1}(x)=Ax+B$ $y_{p1}'(x)=A$ $y_{p1}''(x)=0$ $A=-\frac12$ $B=-1$ $y_{p1}=-\frac12x-1$

1. MathSofiya

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

2. MathSofiya

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

3. MathSofiya

@satellite73

4. satellite73

really i am sorry i have no clue

5. psi9epsilon

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

So what should I do?

7. psi9epsilon

Did you try this using wronskian

8. MathSofiya

I don't know what that is

9. psi9epsilon

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

did you look up wronskian

11. across

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

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

13. MathSofiya

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

14. across

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

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

16. TuringTest

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

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

18. TuringTest

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

19. TuringTest

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

20. across

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

@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

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

23. TuringTest

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

24. TuringTest
25. TuringTest

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

26. across

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

thanks! Gotta get sharp on that technique

28. MathSofiya

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

29. TuringTest

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

30. MathSofiya

correct

31. TuringTest

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

32. MathSofiya

Asin2x+Bcos2x?

33. TuringTest

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

ok

35. TuringTest

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

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

linearly independent*

38. TuringTest

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

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

40. across

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

41. TuringTest

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

42. MathSofiya

A=1

43. TuringTest

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

44. MathSofiya

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

45. MathSofiya

Thanks to you too @across

46. MathSofiya

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

47. MathSofiya

?

48. TuringTest

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

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

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

51. MathSofiya

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

52. TuringTest

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

53. MathSofiya

No hurry =P

54. TuringTest

I did get A=-1/2 by hand

55. TuringTest

actually you can see that fairly quickly...

56. TuringTest

wanna see how?

57. MathSofiya

sure

58. MathSofiya

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

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

okay I have confirmed wolf's answer on paper

61. MathSofiya

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

62. MathSofiya

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

63. TuringTest

you wanna walk through it?

64. MathSofiya

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

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

66. MathSofiya

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

67. TuringTest

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

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

69. TuringTest

yep

70. MathSofiya

something is still wrong though

71. TuringTest

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

72. MathSofiya

oh I see what I did

73. TuringTest

sweet

74. MathSofiya

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

no it aint :P

76. TuringTest

you changed a sign

77. TuringTest

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

78. MathSofiya

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

yay!

80. MathSofiya

LATEX makes all the difference!

81. TuringTest

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

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

83. TuringTest

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

I mean in the complimentary

85. across

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

86. MathSofiya

oh my gosh! did you just make that?

87. across

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

88. TuringTest

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

89. MathSofiya

you are soo sweet! thank you!

90. MathSofiya

OS rocks!

91. TuringTest

yes it does :)