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

Mathematics
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\[x^3y'' + x^2y' -4xy = 1\]\[x^2y'' + xy' -4y = \frac{1}{x}\]\[x^2y_c'' + xy_c' -4y_c =0\]\[\lambda (\lambda-1)+\lambda - 4 =0\]\[\lambda = \pm2\]\[y_c = c_1x^2+c_2x^{-2}\]
hi @RolyPoly are you in DE
\[u_1'(x^2)+u_2'(x^{-2})=0\]\[u_1'(2x)+u_2'(\frac{-2}{x^3})=\frac{1}{x}\] \[u_1'(x^2)+u_2'(x^{-2})=0\]\[u_1'(2x^2)+u_2'(\frac{-2}{x^2})=1\] \[4x^2u_1'=1\]\[u_1'= \frac{1}{4x^2}\]\[u_1 = -\frac{1}{4x}\] I'm doing something wrong, probably.. @mark_o. Yes.

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Other answers:

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
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'')
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
@sirm3d can you give us some input here?
: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
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?
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 :\
\[\large y''=\frac{v''}{x^2}-2\frac{v'}{x^3}-2\frac{v'}{x^3}+6\frac{v}{x^4}\]
Hmm.. Would you mind explaining why you did so? It's, somehow, different from what I've learnt in the lesson..
derivative of a product \(v\) and \(1/x^2\)
I... mean why you did so...
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 ..
\[x^3y''+x^2y'-4xy=x^3\left( \frac{ v'' }{ x^2 }-4\frac{ v' }{ x^3 } +6\frac{ v }{ x^4 }\right)+x^2\left( \frac{ v' }{ x^2 }-\frac{ 2v }{ x^3 }\right)-4x \left( \frac{ v }{ x^2 } \right)\]\[=xv''-4v'+6\frac{ v }{ x }+v'-2\frac{ v }{ x }-4\frac{ v }{ x }\]\[=xv''-3v'=1\]
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.
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.
\[\large xv''-3v'=1\]let \(w=v'\) and \(w' = v''\)\[\large xw'-3w=1\]\[\large w'-\frac{3}{x}w=\frac{1}{x}\]
\[\large w(\frac{ 1 }{ x^3 })=\int\limits \frac{1}{x^4} dx=-(1/3)x^{-3}\]\[\large v'=w=(-1/3)\]\[\large v=(-1/3)x\]\[\large y_p=v(\frac{ 1 }{ x^2 })=-\frac{ 1 }{ 3x }\]
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.
But I think there are two complementary solutions? \(x^2\) and \(x^{-2}\)?!
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
Wouldn't it be extremely weird?! I think I chose both the complementary functions..but you just chose one.. It's weird...
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...
is the particular solution \[\large y_p=-x e^{2x}+(\ln x) x e^{2x}\]
i got that result following your method.
if i use my method, and choose the complementary solution \(e^{2x}\)\[\large y=v e^{2x}\]\[\large y'=v'e^{2x}+2v e^{2x}\]\[\large y''=v'' e^{2x}+4v'e^{2x}+4v e^{2x}\]
\[\large y''-4y' + 4y=(v''e^{2x}+4v'e^{2x}+4v e^{2x})-4(v'e^{2x} + 2v e^{2x})+4(v e^{2x})\]\[\large =v''e^{2x}=\frac{e^{2x}}{x}\]\[\large v''=\frac{1}{x}\]
\[\large v'=\ln x\]\[\large v=x\ln x - x\]\[\large y=\left(x \ln x - x \right) e^{2x}\], no different from the result using your method.
Is.. it...magic? :S Please let me try to do it using your method!!
can you provide me a link on your method?
Link!? I don't have it :( Or can I just type the notes I copied in the lesson here? (scanner is not working now :( )
i found a link detailing your method. i'm reading it now.
@RolyPoly i saw some mistakes in your solution. your two equations are \[\large u_1'x^2+u_2x^{-2}=0\]\[\large u_1(2x)+u_2(-2x^{-3})=\frac{1/x}{x^2}=\frac{1}{x^3}\]
Why should it be (1/x)/x^2 ??
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 \(-1/(3x)\)
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 \[\large b_nx^n y^{(n)}+\cdots +b_1xy'+b_0y=r(x)\]
you should get \(y_p=-1/(3x)\) this time after solving \(u_1\) and \(u_2\). this are the values i got: \[\large u_1=-\frac{1}{12x^3}\]\[\large u_2=-\frac{x}{4}\]
@RolyPoly hope i have given you peace of mind in this problem.
My... teacher... didn't .... told... us... that... the ...second equation is in the form \[ u_1'y_1'+u_2'y_2'=\frac{r(x)}{b_nx^n}\]
that may be the cause of your problems.
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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