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

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RolyPolyBest ResponseYou've already chosen the best response.0
\[x^3y'' + x^2y' 4xy = 1\]\[x^2y'' + xy' 4y = \frac{1}{x}\]\[x^2y_c'' + xy_c' 4y_c =0\]\[\lambda (\lambda1)+\lambda  4 =0\]\[\lambda = \pm2\]\[y_c = c_1x^2+c_2x^{2}\]
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
hi @RolyPoly are you in DE
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
\[u_1'(x^2)+u_2'(x^{2})=0\]\[u_1'(2x)+u_2'(\frac{2}{x^3})=\frac{1}{x}\] \[u_1'(x^2)+u_2'(x^{2})=0\]\[u_1'(2x^2)+u_2'(\frac{2}{x^2})=1\] \[4x^2u_1'=1\]\[u_1'= \frac{1}{4x^2}\]\[u_1 = \frac{1}{4x}\] I'm doing something wrong, probably.. @mark_o. Yes.
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
are you familiar with cauchy euler equation? that is the problem you are having there
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
are your class in that chapter already?
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
chapter in cauchy euler equation ?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Yes. I'm having trouble with finding the particular solutions.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
The CauchyEuler part is done, I think.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
I divided both sides by x...
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
And isn't it a second order DE only?
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
thats the formula for getting the characteristic equation
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Why doesn't my method work?
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
hold on lets double check it by using this formula
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
And for CauchyEuler equation, isn't it the power of x and the no. of times of derivatives of y the same? (eg: x^2y'')
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
it may not be the same so lets try doing the the formula one im giving you
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
But this is not a 3rd order DE!
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
im sorry i mess up,, yes its a 2nd order
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
So will that m^3 thing work here?
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
no,we have to use the (am^2+(ba)m+c)x^m=0
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Wait.. Please refer to my first comment here. The CauchyEuler thing is done, I think.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
The problem is how to find the particular solution.
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
so that we can reduced i/x into function of t
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
and it's y=...(in terms of x) How come we have a t?!
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
let x=e^t then 1/x==1/e^t=e^t
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
temporarily so that we can get Yp
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
It's not the form of yp
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
Y=ae^t Y'=Ae^t Y''=Ae^t now sub this to your equation
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
hmm try reading this first http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
I think you need to know this is not a CauchyEuler equation.
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
http://www.sosmath.com/diffeq/second/euler/euler.html
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Since the right side \(\ne\) 0
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
eventhough its not zero you can solve it using cauchy  euler eq
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
this may be a good one to read http://www.sosmath.com/diffeq/second/euler/euler.html
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
But CauchyEuler is over after we get the yc, isn't it?
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
not until you convert everything to x=e^t
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
try to convert the whole thing to x=e^t
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
so let a=1, b=i,c=4 then Y''(t)+0Y'(t)4Y=e^t that will be your equation
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
I'm sorry, I must go now. Can we continue tomorrow?
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
m^2_4=0 m=+2.2 Yc(t)=C1e^2t +C2e^2 next finding Yp
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
that was m^24=0 m=2,+2
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
yes ok we can continue tomorrow... :D
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
ok YW good luck now and have fun :D
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
@sirm3d can you give us some input here?
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
.......................... :D
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
by variation of parameters, let \(y=v\cdot\frac{1}{x^2}\)
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
its been awhile since i took DE, i need to brush up again lol
 one year ago

mark_o.Best ResponseYou've already chosen the best response.0
yesss variation of parameter too...but i think euler is much easier
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
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?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
differentiation of product yields \[\large y'=\frac{v'}{x^2}2\frac{v}{x^3}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
But shouldn't we ''kill'' the last term to avoid troubles when differentiate y' again?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Sorry, I meant the first term :\
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[\large y''=\frac{v''}{x^2}2\frac{v'}{x^3}2\frac{v'}{x^3}+6\frac{v}{x^4}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Hmm.. Would you mind explaining why you did so? It's, somehow, different from what I've learnt in the lesson..
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
derivative of a product \(v\) and \(1/x^2\)
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
I... mean why you did so...
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
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 ..
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[x^3y''+x^2y'4xy=x^3\left( \frac{ v'' }{ x^2 }4\frac{ v' }{ x^3 } +6\frac{ v }{ x^4 }\right)+x^2\left( \frac{ v' }{ x^2 }\frac{ 2v }{ x^3 }\right)4x \left( \frac{ v }{ x^2 } \right)\]\[=xv''4v'+6\frac{ v }{ x }+v'2\frac{ v }{ x }4\frac{ v }{ x }\]\[=xv''3v'=1\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.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 cauchyeuler method.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
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.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[\large xv''3v'=1\]let \(w=v'\) and \(w' = v''\)\[\large xw'3w=1\]\[\large w'\frac{3}{x}w=\frac{1}{x}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[\large w(\frac{ 1 }{ x^3 })=\int\limits \frac{1}{x^4} dx=(1/3)x^{3}\]\[\large v'=w=(1/3)\]\[\large v=(1/3)x\]\[\large y_p=v(\frac{ 1 }{ x^2 })=\frac{ 1 }{ 3x }\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
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.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
But I think there are two complementary solutions? \(x^2\) and \(x^{2}\)?!
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
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
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Wouldn't it be extremely weird?! I think I chose both the complementary functions..but you just chose one.. It's weird...
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
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...
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
is the particular solution \[\large y_p=x e^{2x}+(\ln x) x e^{2x}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
i got that result following your method.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
if i use my method, and choose the complementary solution \(e^{2x}\)\[\large y=v e^{2x}\]\[\large y'=v'e^{2x}+2v e^{2x}\]\[\large y''=v'' e^{2x}+4v'e^{2x}+4v e^{2x}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[\large y''4y' + 4y=(v''e^{2x}+4v'e^{2x}+4v e^{2x})4(v'e^{2x} + 2v e^{2x})+4(v e^{2x})\]\[\large =v''e^{2x}=\frac{e^{2x}}{x}\]\[\large v''=\frac{1}{x}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[\large v'=\ln x\]\[\large v=x\ln x  x\]\[\large y=\left(x \ln x  x \right) e^{2x}\], no different from the result using your method.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Is.. it...magic? :S Please let me try to do it using your method!!
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
can you provide me a link on your method?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Link!? I don't have it :( Or can I just type the notes I copied in the lesson here? (scanner is not working now :( )
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
i found a link detailing your method. i'm reading it now.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
@RolyPoly i saw some mistakes in your solution. your two equations are \[\large u_1'x^2+u_2x^{2}=0\]\[\large u_1(2x)+u_2(2x^{3})=\frac{1/x}{x^2}=\frac{1}{x^3}\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Why should it be (1/x)/x^2 ??
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
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 \(1/(3x)\)
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
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 cauchyeuler \[\large b_nx^n y^{(n)}+\cdots +b_1xy'+b_0y=r(x)\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
you should get \(y_p=1/(3x)\) this time after solving \(u_1\) and \(u_2\). this are the values i got: \[\large u_1=\frac{1}{12x^3}\]\[\large u_2=\frac{x}{4}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
@RolyPoly hope i have given you peace of mind in this problem.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
My... teacher... didn't .... told... us... that... the ...second equation is in the form \[ u_1'y_1'+u_2'y_2'=\frac{r(x)}{b_nx^n}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
that may be the cause of your problems.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
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!
 one year ago
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