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

1. RolyPoly

$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. mark_o.

hi @RolyPoly are you in DE

3. RolyPoly

$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. mark_o.

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

5. mark_o.

6. mark_o.

chapter in cauchy -euler equation ?

7. RolyPoly

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

8. RolyPoly

The Cauchy-Euler part is done, I think.

9. RolyPoly

I divided both sides by x...

10. RolyPoly

And isn't it a second order DE only?

11. RolyPoly

Wwhat is it??

12. mark_o.

thats the formula for getting the characteristic equation

13. RolyPoly

Why doesn't my method work?

14. mark_o.

hold on lets double check it by using this formula

15. RolyPoly

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. mark_o.

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

17. RolyPoly

But this is not a 3rd order DE!

18. mark_o.

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

19. RolyPoly

So will that m^3 thing work here?

20. mark_o.

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

21. RolyPoly

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

22. mark_o.

23. RolyPoly

The problem is how to find the particular solution.

24. mark_o.

ok take x=e^t

25. RolyPoly

Why?

26. mark_o.

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

27. RolyPoly

Not i/x but 1/x

28. mark_o.

yes 1/x

29. RolyPoly

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

30. mark_o.

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

31. mark_o.

temporarily so that we can get Yp

32. RolyPoly

How come...

33. mark_o.

assume Yp=Ae^-t

34. RolyPoly

It's not the form of yp

35. mark_o.

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

36. RolyPoly

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

37. mark_o.

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

38. RolyPoly

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

39. mark_o.
40. RolyPoly

Since the right side $$\ne$$ 0

41. mark_o.

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

42. mark_o.

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

43. RolyPoly

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

44. mark_o.

not until you convert everything to x=e^t

45. mark_o.

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

46. mark_o.

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

47. RolyPoly

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

48. mark_o.

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

49. mark_o.

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

50. mark_o.

yes ok we can continue tomorrow... :D

51. RolyPoly

Thanks.

52. mark_o.

ok YW good luck now and have fun :D

53. mark_o.

@sirm3d can you give us some input here?

54. mark_o.

:D

55. mark_o.

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

56. sirm3d

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

57. mark_o.

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

58. mark_o.

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

59. RolyPoly

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. sirm3d

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

61. RolyPoly

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

62. RolyPoly

Sorry, I meant the first term :\

63. sirm3d

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

64. RolyPoly

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

65. sirm3d

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

66. RolyPoly

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

67. RolyPoly

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. sirm3d

$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. sirm3d

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. sirm3d

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. sirm3d

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

72. sirm3d

$\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. sirm3d

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. RolyPoly

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

75. sirm3d

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. RolyPoly

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

77. RolyPoly

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. sirm3d

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

79. sirm3d

i got that result following your method.

80. sirm3d

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. sirm3d

$\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. sirm3d

$\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. RolyPoly

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

84. sirm3d

85. RolyPoly

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. sirm3d

87. sirm3d

@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. RolyPoly

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

89. sirm3d

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. sirm3d

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. sirm3d

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. sirm3d

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

93. RolyPoly

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. sirm3d

that may be the cause of your problems.

95. RolyPoly

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!