anonymous
  • anonymous
Solve by undetermined coefficients/variation of parameters: \(x^3y'' + x^2y' -4xy = 1\) , x>0
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • 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
  • anonymous
hi @RolyPoly are you in DE
anonymous
  • 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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anonymous
  • anonymous
are you familiar with cauchy -euler equation? that is the problem you are having there
anonymous
  • anonymous
are your class in that chapter already?
anonymous
  • anonymous
chapter in cauchy -euler equation ?
anonymous
  • anonymous
Yes. I'm having trouble with finding the particular solutions.
anonymous
  • anonymous
The Cauchy-Euler part is done, I think.
anonymous
  • anonymous
I divided both sides by x...
anonymous
  • anonymous
And isn't it a second order DE only?
anonymous
  • anonymous
Wwhat is it??
anonymous
  • anonymous
thats the formula for getting the characteristic equation
anonymous
  • anonymous
Why doesn't my method work?
anonymous
  • anonymous
hold on lets double check it by using this formula
anonymous
  • 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
  • anonymous
it may not be the same so lets try doing the the formula one im giving you
anonymous
  • anonymous
But this is not a 3rd order DE!
anonymous
  • anonymous
im sorry i mess up,, yes its a 2nd order
anonymous
  • anonymous
So will that m^3 thing work here?
anonymous
  • anonymous
no,we have to use the (am^2+(b-a)m+c)x^m=0
anonymous
  • anonymous
Wait.. Please refer to my first comment here. The Cauchy-Euler thing is done, I think.
anonymous
  • anonymous
yes i think your ok
anonymous
  • anonymous
The problem is how to find the particular solution.
anonymous
  • anonymous
ok take x=e^t
anonymous
  • anonymous
Why?
anonymous
  • anonymous
so that we can reduced i/x into function of t
anonymous
  • anonymous
Not i/x but 1/x
anonymous
  • anonymous
yes 1/x
anonymous
  • anonymous
and it's y=...(in terms of x) How come we have a t?!
anonymous
  • anonymous
let x=e^t then 1/x==1/e^t=e^-t
anonymous
  • anonymous
temporarily so that we can get Yp
anonymous
  • anonymous
How come...
anonymous
  • anonymous
assume Yp=Ae^-t
anonymous
  • anonymous
It's not the form of yp
anonymous
  • anonymous
Y=ae^-t Y'=-Ae^t Y''=Ae^t now sub this to your equation
anonymous
  • anonymous
The yp is -1/(3x)...
anonymous
  • anonymous
hmm try reading this first http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation
anonymous
  • anonymous
I think you need to know this is not a Cauchy-Euler equation.
anonymous
  • anonymous
http://www.sosmath.com/diffeq/second/euler/euler.html
anonymous
  • anonymous
Since the right side \(\ne\) 0
anonymous
  • anonymous
eventhough its not zero you can solve it using cauchy - euler eq
anonymous
  • anonymous
this may be a good one to read http://www.sosmath.com/diffeq/second/euler/euler.html
anonymous
  • anonymous
But Cauchy-Euler is over after we get the yc, isn't it?
anonymous
  • anonymous
not until you convert everything to x=e^t
anonymous
  • anonymous
try to convert the whole thing to x=e^t
anonymous
  • anonymous
so let a=1, b=i,c=-4 then Y''(t)+0Y'(t)-4Y=e^-t that will be your equation
anonymous
  • anonymous
I'm sorry, I must go now. Can we continue tomorrow?
anonymous
  • anonymous
m^2_4=0 m=+2.-2 Yc(t)=C1e^-2t +C2e^2 next finding Yp
anonymous
  • anonymous
that was m^2-4=0 m=-2,+2
anonymous
  • anonymous
yes ok we can continue tomorrow... :D
anonymous
  • anonymous
Thanks.
anonymous
  • anonymous
ok YW good luck now and have fun :D
anonymous
  • anonymous
@sirm3d can you give us some input here?
anonymous
  • anonymous
:D
anonymous
  • anonymous
.......................... :D
sirm3d
  • sirm3d
by variation of parameters, let \(y=v\cdot\frac{1}{x^2}\)
anonymous
  • anonymous
its been awhile since i took DE, i need to brush up again lol
anonymous
  • anonymous
yesss variation of parameter too...but i think euler is much easier
anonymous
  • 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
  • sirm3d
differentiation of product yields \[\large y'=\frac{v'}{x^2}-2\frac{v}{x^3}\]
anonymous
  • anonymous
But shouldn't we ''kill'' the last term to avoid troubles when differentiate y' again?
anonymous
  • anonymous
Sorry, I meant the first term :\
sirm3d
  • sirm3d
\[\large y''=\frac{v''}{x^2}-2\frac{v'}{x^3}-2\frac{v'}{x^3}+6\frac{v}{x^4}\]
anonymous
  • anonymous
Hmm.. Would you mind explaining why you did so? It's, somehow, different from what I've learnt in the lesson..
sirm3d
  • sirm3d
derivative of a product \(v\) and \(1/x^2\)
anonymous
  • anonymous
I... mean why you did so...
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
But I think there are two complementary solutions? \(x^2\) and \(x^{-2}\)?!
sirm3d
  • 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
  • anonymous
Wouldn't it be extremely weird?! I think I chose both the complementary functions..but you just chose one.. It's weird...
anonymous
  • 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
  • sirm3d
is the particular solution \[\large y_p=-x e^{2x}+(\ln x) x e^{2x}\]
sirm3d
  • sirm3d
i got that result following your method.
sirm3d
  • 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
  • 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
  • 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
  • anonymous
Is.. it...magic? :S Please let me try to do it using your method!!
sirm3d
  • sirm3d
can you provide me a link on your method?
anonymous
  • 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
  • sirm3d
i found a link detailing your method. i'm reading it now.
sirm3d
  • 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
  • anonymous
Why should it be (1/x)/x^2 ??
sirm3d
  • 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
  • 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
  • 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
  • sirm3d
@RolyPoly hope i have given you peace of mind in this problem.
anonymous
  • 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
  • sirm3d
that may be the cause of your problems.
anonymous
  • 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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