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

- anonymous

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

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- anonymous

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

- anonymous

hi @RolyPoly are you in DE

- 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.

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## More answers

- anonymous

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

- anonymous

are your class in that chapter already?

- anonymous

chapter in cauchy -euler equation ?

- anonymous

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

- anonymous

The Cauchy-Euler part is done, I think.

- anonymous

I divided both sides by x...

- anonymous

And isn't it a second order DE only?

- anonymous

Wwhat is it??

- anonymous

thats the formula for getting the characteristic equation

- anonymous

Why doesn't my method work?

- anonymous

hold on lets double check it by using this formula

- 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'')

- anonymous

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

- anonymous

But this is not a 3rd order DE!

- anonymous

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

- anonymous

So will that m^3 thing work here?

- anonymous

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

- anonymous

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

- anonymous

yes i think your ok

- anonymous

The problem is how to find the particular solution.

- anonymous

ok take x=e^t

- anonymous

Why?

- anonymous

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

- anonymous

Not i/x but 1/x

- anonymous

yes 1/x

- anonymous

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

- anonymous

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

- anonymous

temporarily so that we can get Yp

- anonymous

How come...

- anonymous

assume
Yp=Ae^-t

- anonymous

It's not the form of yp

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Since the right side \(\ne\) 0

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

not until you convert everything to x=e^t

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes ok we can continue tomorrow... :D

- anonymous

Thanks.

- anonymous

ok YW good luck now and have fun :D

- anonymous

@sirm3d can you give us some input here?

- anonymous

:D

- anonymous

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

- sirm3d

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

- anonymous

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

- anonymous

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

- 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?

- sirm3d

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

- anonymous

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

- anonymous

Sorry, I meant the first term :\

- sirm3d

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

- anonymous

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

- sirm3d

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

- anonymous

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

- 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 ..

- sirm3d

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

- 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.

- 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.

- sirm3d

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

- sirm3d

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

- 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.

- anonymous

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

- 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

- anonymous

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

- 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...

- sirm3d

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

- sirm3d

i got that result following your method.

- sirm3d

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

- sirm3d

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

- 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.

- anonymous

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

- sirm3d

can you provide me a link on your method?

- 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 :( )

- sirm3d

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

- sirm3d

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

- anonymous

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

- 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 \(-1/(3x)\)

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

- sirm3d

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

- sirm3d

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

- anonymous

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

- sirm3d

that may be the cause of your problems.

- 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!

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