Question

Solve by undetermined coefficients/variation of parameters:
\(x^3y'' + x^2y' -4xy = 1\) , x>0

Answer 1

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

Answer 2

hi @RolyPoly are you in DE

Answer 3

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

Answer 4

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

Answer 5

are your class in that chapter already?

Answer 6

chapter in cauchy -euler equation ?

Answer 7

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

Answer 8

The Cauchy-Euler part is done, I think.

Answer 9

I divided both sides by x...

Answer 10

And isn't it a second order DE only?

Answer 11

Wwhat is it??

Answer 12

thats the formula for getting the characteristic equation

Answer 13

Why doesn't my method work?

Answer 14

hold on lets double check it by using this formula

Answer 15

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'')

Answer 16

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

Answer 17

But this is not a 3rd order DE!

Answer 18

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

Answer 19

So will that m^3 thing work here?

Answer 20

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

Answer 21

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

Answer 22

yes i think your ok

Answer 23

The problem is how to find the particular solution.

Answer 24

ok take x=e^t

Answer 25

Why?

Answer 26

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

Answer 27

Not i/x but 1/x

Answer 28

yes 1/x

Answer 29

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

Answer 30

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

Answer 31

temporarily so that we can get Yp

Answer 32

How come...

Answer 33

assume
Yp=Ae^-t

Answer 34

It's not the form of yp

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

http://www.sosmath.com/diffeq/second/euler/euler.html

Answer 40

Since the right side \(\ne\) 0

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

not until you convert everything to x=e^t

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

yes ok we can continue tomorrow... :D

Answer 51

Thanks.

Answer 52

ok YW good luck now and have fun :D

Answer 53

@sirm3d can you give us some input here?

Answer 54

:D

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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?

Answer 60

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

Answer 61

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

Answer 62

Sorry, I meant the first term :\

Answer 63

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

Answer 64

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

Answer 65

derivative of a product \(v\) and \(1/x^2\)

Answer 66

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

Answer 67

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 ..

Answer 68

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

Answer 69

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.

Answer 70

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.

Answer 71

\[\large xv''-3v'=1\]let \(w=v'\) and \(w' = v''\)\[\large xw'-3w=1\]\[\large w'-\frac{3}{x}w=\frac{1}{x}\]

Answer 72

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

Answer 73

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.

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

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...

Answer 78

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

Answer 79

i got that result following your method.

Answer 80

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

Answer 81

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

Answer 82

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

Answer 83

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

Answer 84

can you provide me a link on your method?

Answer 85

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

Answer 86

i found a link detailing your method. i'm reading it now.

Answer 87

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

Answer 88

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

Answer 89

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 \(-1/(3x)\)

Answer 90

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 \[\large b_nx^n y^{(n)}+\cdots +b_1xy'+b_0y=r(x)\]

Answer 91

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

Answer 92

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

Answer 93

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

Answer 94

that may be the cause of your problems.

Answer 95

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!