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

are your class in that chapter already?

chapter in cauchy -euler equation ?

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

The Cauchy-Euler part is done, I think.

I divided both sides by x...

And isn't it a second order DE only?

Wwhat is it??

thats the formula for getting the characteristic equation

Why doesn't my method work?

hold on lets double check it by using this formula

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

But this is not a 3rd order DE!

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

So will that m^3 thing work here?

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

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

yes i think your ok

The problem is how to find the particular solution.

ok take x=e^t

Why?

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

Not i/x but 1/x

yes 1/x

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

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

temporarily so that we can get Yp

How come...

assume
Yp=Ae^-t

It's not the form of yp

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

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

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

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

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

Since the right side \(\ne\) 0

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

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

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

not until you convert everything to x=e^t

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

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

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

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

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

yes ok we can continue tomorrow... :D

Thanks.

ok YW good luck now and have fun :D

:D

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

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

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

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

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

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

Sorry, I meant the first term :\

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

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

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

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

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

i got that result following your method.

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

can you provide me a link on your method?

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

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

that may be the cause of your problems.