## anonymous 3 years ago Solve by undetermined coefficients/variation of parameters: $$x^3y'' + x^2y' -4xy = 1$$ , x>0

1. anonymous

$x^3y'' + x^2y' -4xy = 1$$x^2y'' + xy' -4y = \frac{1}{x}$$x^2y_c'' + xy_c' -4y_c =0$$\lambda (\lambda-1)+\lambda - 4 =0$$\lambda = \pm2$$y_c = c_1x^2+c_2x^{-2}$

2. anonymous

hi @RolyPoly are you in DE

3. anonymous

$u_1'(x^2)+u_2'(x^{-2})=0$$u_1'(2x)+u_2'(\frac{-2}{x^3})=\frac{1}{x}$ $u_1'(x^2)+u_2'(x^{-2})=0$$u_1'(2x^2)+u_2'(\frac{-2}{x^2})=1$ $4x^2u_1'=1$$u_1'= \frac{1}{4x^2}$$u_1 = -\frac{1}{4x}$ I'm doing something wrong, probably.. @mark_o. Yes.

4. anonymous

are you familiar with cauchy -euler equation? that is the problem you are having there

5. anonymous

6. anonymous

chapter in cauchy -euler equation ?

7. anonymous

Yes. I'm having trouble with finding the particular solutions.

8. anonymous

The Cauchy-Euler part is done, I think.

9. anonymous

I divided both sides by x...

10. anonymous

And isn't it a second order DE only?

11. anonymous

Wwhat is it??

12. anonymous

thats the formula for getting the characteristic equation

13. anonymous

Why doesn't my method work?

14. anonymous

hold on lets double check it by using this formula

15. anonymous

And for Cauchy-Euler equation, isn't it the power of x and the no. of times of derivatives of y the same? (eg: x^2y'')

16. anonymous

it may not be the same so lets try doing the the formula one im giving you

17. anonymous

But this is not a 3rd order DE!

18. anonymous

im sorry i mess up,, yes its a 2nd order

19. anonymous

So will that m^3 thing work here?

20. anonymous

no,we have to use the (am^2+(b-a)m+c)x^m=0

21. anonymous

Wait.. Please refer to my first comment here. The Cauchy-Euler thing is done, I think.

22. anonymous

23. anonymous

The problem is how to find the particular solution.

24. anonymous

ok take x=e^t

25. anonymous

Why?

26. anonymous

so that we can reduced i/x into function of t

27. anonymous

Not i/x but 1/x

28. anonymous

yes 1/x

29. anonymous

and it's y=...(in terms of x) How come we have a t?!

30. anonymous

let x=e^t then 1/x==1/e^t=e^-t

31. anonymous

temporarily so that we can get Yp

32. anonymous

How come...

33. anonymous

assume Yp=Ae^-t

34. anonymous

It's not the form of yp

35. anonymous

Y=ae^-t Y'=-Ae^t Y''=Ae^t now sub this to your equation

36. anonymous

The yp is -1/(3x)...

37. anonymous

hmm try reading this first http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation

38. anonymous

I think you need to know this is not a Cauchy-Euler equation.

39. anonymous
40. anonymous

Since the right side $$\ne$$ 0

41. anonymous

eventhough its not zero you can solve it using cauchy - euler eq

42. anonymous

this may be a good one to read http://www.sosmath.com/diffeq/second/euler/euler.html

43. anonymous

But Cauchy-Euler is over after we get the yc, isn't it?

44. anonymous

not until you convert everything to x=e^t

45. anonymous

try to convert the whole thing to x=e^t

46. anonymous

so let a=1, b=i,c=-4 then Y''(t)+0Y'(t)-4Y=e^-t that will be your equation

47. anonymous

I'm sorry, I must go now. Can we continue tomorrow?

48. anonymous

m^2_4=0 m=+2.-2 Yc(t)=C1e^-2t +C2e^2 next finding Yp

49. anonymous

that was m^2-4=0 m=-2,+2

50. anonymous

yes ok we can continue tomorrow... :D

51. anonymous

Thanks.

52. anonymous

ok YW good luck now and have fun :D

53. anonymous

@sirm3d can you give us some input here?

54. anonymous

:D

55. anonymous

.......................... :D

56. anonymous

by variation of parameters, let $$y=v\cdot\frac{1}{x^2}$$

57. anonymous

its been awhile since i took DE, i need to brush up again lol

58. anonymous

yesss variation of parameter too...but i think euler is much easier

59. anonymous

Looks like I'm not doing this right... $y=\frac{v}{x^2}$$y'=-\frac{2v}{x^3}$$y''=-\frac{2v'}{x^3}+\frac{6v}{x^4}$ $x^3(-\frac{2v'}{x^3}+\frac{6v}{x^4}) +x^2(-\frac{2v}{x^3}) - 4x(\frac{v}{x^2})=1$$-2v'+\frac{6v}{x} -\frac{2v}{x} - \frac{4v}{x}=1$$v'=-\frac{1}{2}$$v=-\frac{1}{2}x$ $y_p=-\frac{1}{2x}$Hmm... Where did I make mistakes?

60. anonymous

differentiation of product yields $\large y'=\frac{v'}{x^2}-2\frac{v}{x^3}$

61. anonymous

But shouldn't we ''kill'' the last term to avoid troubles when differentiate y' again?

62. anonymous

Sorry, I meant the first term :\

63. anonymous

$\large y''=\frac{v''}{x^2}-2\frac{v'}{x^3}-2\frac{v'}{x^3}+6\frac{v}{x^4}$

64. anonymous

Hmm.. Would you mind explaining why you did so? It's, somehow, different from what I've learnt in the lesson..

65. anonymous

derivative of a product $$v$$ and $$1/x^2$$

66. anonymous

I... mean why you did so...

67. anonymous

When way I've learnt is to let $$u_1y_1$$, $$u_2y_2$$be the particular solution.. But somehow.. you're doing something different (in this question and in the previous one). I really want to know why you did that and how you came up with this y=v/x^2 ..

68. anonymous

$x^3y''+x^2y'-4xy=x^3\left( \frac{ v'' }{ x^2 }-4\frac{ v' }{ x^3 } +6\frac{ v }{ x^4 }\right)+x^2\left( \frac{ v' }{ x^2 }-\frac{ 2v }{ x^3 }\right)-4x \left( \frac{ v }{ x^2 } \right)$$=xv''-4v'+6\frac{ v }{ x }+v'-2\frac{ v }{ x }-4\frac{ v }{ x }$$=xv''-3v'=1$

69. anonymous

the idea behind the variation of parameter is to use one complementary function that is a solution of the homogeneous DE. In this problem, you produced one complementary function$$1/x^2$$ using cauchy-euler method.

70. anonymous

the variation of parameter introduces a function $$v$$ to solve THE particular solution using one complementary solution $$1/x^2$$ by assuming that $$y_p=v(1/x^2)$$ is the particular solution of the DE, where v is to be determined.

71. anonymous

$\large xv''-3v'=1$let $$w=v'$$ and $$w' = v''$$$\large xw'-3w=1$$\large w'-\frac{3}{x}w=\frac{1}{x}$

72. anonymous

$\large w(\frac{ 1 }{ x^3 })=\int\limits \frac{1}{x^4} dx=-(1/3)x^{-3}$$\large v'=w=(-1/3)$$\large v=(-1/3)x$$\large y_p=v(\frac{ 1 }{ x^2 })=-\frac{ 1 }{ 3x }$

73. anonymous

check: $\large x^3y''+x^2y'-4xy=x^3\left( -\frac{ 2 }{ 3x^3 } \right)+x^2\left( \frac{ 1 }{ 3x^2 } \right)-4x \left( -\frac{ 1 }{ 3x } \right)$$\large -\frac{ 2 }{ 3 }+\frac{ 1 }{ 3 }+\frac{ 4 }{ 3 }=1$ and so it's solved.

74. anonymous

But I think there are two complementary solutions? $$x^2$$ and $$x^{-2}$$?!

75. anonymous

variation of parameter allows you to choose any of the complementary solutions. i chose $$x^{-2}$$ over $$x^2$$ for no particular reason. why don't you try $$y=vx^2$$ ? i have yet to learn your method to find why you get -1/4 when i got -1/3

76. anonymous

Wouldn't it be extremely weird?! I think I chose both the complementary functions..but you just chose one.. It's weird...

77. anonymous

Consider y''-4y'+4y = e^(2x) / x $y_c = c_1e^{2x} +c_2xe^{2x}$ The way I find the particular solution is $u_1'e^{2x}+u_2'(xe^{2x})=0$$u_1'(2e^{2x})+u_2'(2xe^{2x}+e^{2x})=\frac{e^{2x}}{x}$and find u1 u2. Then, there are two parts in the particular solution...

78. anonymous

is the particular solution $\large y_p=-x e^{2x}+(\ln x) x e^{2x}$

79. anonymous

i got that result following your method.

80. anonymous

if i use my method, and choose the complementary solution $$e^{2x}$$$\large y=v e^{2x}$$\large y'=v'e^{2x}+2v e^{2x}$$\large y''=v'' e^{2x}+4v'e^{2x}+4v e^{2x}$

81. anonymous

$\large y''-4y' + 4y=(v''e^{2x}+4v'e^{2x}+4v e^{2x})-4(v'e^{2x} + 2v e^{2x})+4(v e^{2x})$$\large =v''e^{2x}=\frac{e^{2x}}{x}$$\large v''=\frac{1}{x}$

82. anonymous

$\large v'=\ln x$$\large v=x\ln x - x$$\large y=\left(x \ln x - x \right) e^{2x}$, no different from the result using your method.

83. anonymous

Is.. it...magic? :S Please let me try to do it using your method!!

84. anonymous

85. anonymous

Link!? I don't have it :( Or can I just type the notes I copied in the lesson here? (scanner is not working now :( )

86. anonymous

87. anonymous

@RolyPoly i saw some mistakes in your solution. your two equations are $\large u_1'x^2+u_2x^{-2}=0$$\large u_1(2x)+u_2(-2x^{-3})=\frac{1/x}{x^2}=\frac{1}{x^3}$

88. anonymous

Why should it be (1/x)/x^2 ??

89. anonymous

just solve $$u_1$$ and $$u_2$$ and sub them in $\large y_p= u_1x^2+u_2x^{-2}$ and you should get the same answer $$-1/(3x)$$

90. anonymous

the second equation in the system is of the form $\large u_1'y_1'+u_2'y_2'=\frac{r(x)}{b_nx^n}$ from the cauchy-euler $\large b_nx^n y^{(n)}+\cdots +b_1xy'+b_0y=r(x)$

91. anonymous

you should get $$y_p=-1/(3x)$$ this time after solving $$u_1$$ and $$u_2$$. this are the values i got: $\large u_1=-\frac{1}{12x^3}$$\large u_2=-\frac{x}{4}$

92. anonymous

@RolyPoly hope i have given you peace of mind in this problem.

93. anonymous

My... teacher... didn't .... told... us... that... the ...second equation is in the form $u_1'y_1'+u_2'y_2'=\frac{r(x)}{b_nx^n}$

94. anonymous

that may be the cause of your problems.

95. anonymous

Most probably.. That seems your method works better.. I have to try both methods to decide which to use.. Thanks!!! I hope you don't mind me tagging you if I still have problems concerning this question!