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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hi @RolyPoly are you in DE

anonymous
 4 years ago
Best 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.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are your class in that chapter already?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0chapter in cauchy euler equation ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. I'm having trouble with finding the particular solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The CauchyEuler part is done, I think.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I divided both sides by x...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And isn't it a second order DE only?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats the formula for getting the characteristic equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why doesn't my method work?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hold on lets double check it by using this formula

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But this is not a 3rd order DE!

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So will that m^3 thing work here?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait.. Please refer to my first comment here. The CauchyEuler thing is done, I think.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The problem is how to find the particular solution.

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let x=e^t then 1/x==1/e^t=e^t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0temporarily so that we can get Yp

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's not the form of yp

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think you need to know this is not a CauchyEuler equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since the right side \(\ne\) 0

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But CauchyEuler is over after we get the yc, isn't it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not until you convert everything to x=e^t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0try to convert the whole thing to x=e^t

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, I must go now. Can we continue tomorrow?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that was m^24=0 m=2,+2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes ok we can continue tomorrow... :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok YW good luck now and have fun :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@sirm3d can you give us some input here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0.......................... :D

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1by variation of parameters, let \(y=v\cdot\frac{1}{x^2}\)

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Looks 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
 4 years ago
Best ResponseYou've already chosen the best response.1differentiation of product yields \[\large y'=\frac{v'}{x^2}2\frac{v}{x^3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But shouldn't we ''kill'' the last term to avoid troubles when differentiate y' again?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, I meant the first term :\

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1\[\large y''=\frac{v''}{x^2}2\frac{v'}{x^3}2\frac{v'}{x^3}+6\frac{v}{x^4}\]

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

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1derivative of a product \(v\) and \(1/x^2\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I... mean why you did so...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When 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
 4 years ago
Best ResponseYou've already chosen the best response.1\[x^3y''+x^2y'4xy=x^3\left( \frac{ v'' }{ x^2 }4\frac{ v' }{ x^3 } +6\frac{ v }{ x^4 }\right)+x^2\left( \frac{ v' }{ x^2 }\frac{ 2v }{ x^3 }\right)4x \left( \frac{ v }{ x^2 } \right)\]\[=xv''4v'+6\frac{ v }{ x }+v'2\frac{ v }{ x }4\frac{ v }{ x }\]\[=xv''3v'=1\]

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1the 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.

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1the 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
 4 years ago
Best ResponseYou've already chosen the best response.1\[\large xv''3v'=1\]let \(w=v'\) and \(w' = v''\)\[\large xw'3w=1\]\[\large w'\frac{3}{x}w=\frac{1}{x}\]

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1\[\large w(\frac{ 1 }{ x^3 })=\int\limits \frac{1}{x^4} dx=(1/3)x^{3}\]\[\large v'=w=(1/3)\]\[\large v=(1/3)x\]\[\large y_p=v(\frac{ 1 }{ x^2 })=\frac{ 1 }{ 3x }\]

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1check: \[\large x^3y''+x^2y'4xy=x^3\left( \frac{ 2 }{ 3x^3 } \right)+x^2\left( \frac{ 1 }{ 3x^2 } \right)4x \left( \frac{ 1 }{ 3x } \right)\]\[\large \frac{ 2 }{ 3 }+\frac{ 1 }{ 3 }+\frac{ 4 }{ 3 }=1\] and so it's solved.

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

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1variation 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
 4 years ago
Best ResponseYou've already chosen the best response.0Wouldn't it be extremely weird?! I think I chose both the complementary functions..but you just chose one.. It's weird...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Consider 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
 4 years ago
Best ResponseYou've already chosen the best response.1is the particular solution \[\large y_p=x e^{2x}+(\ln x) x e^{2x}\]

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1i got that result following your method.

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1if 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
 4 years ago
Best ResponseYou've already chosen the best response.1\[\large y''4y' + 4y=(v''e^{2x}+4v'e^{2x}+4v e^{2x})4(v'e^{2x} + 2v e^{2x})+4(v e^{2x})\]\[\large =v''e^{2x}=\frac{e^{2x}}{x}\]\[\large v''=\frac{1}{x}\]

sirm3d
 4 years ago
Best 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is.. it...magic? :S Please let me try to do it using your method!!

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1can you provide me a link on your method?

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

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1i found a link detailing your method. i'm reading it now.

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1@RolyPoly i saw some mistakes in your solution. your two equations are \[\large u_1'x^2+u_2x^{2}=0\]\[\large u_1(2x)+u_2(2x^{3})=\frac{1/x}{x^2}=\frac{1}{x^3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why should it be (1/x)/x^2 ??

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1just 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
 4 years ago
Best ResponseYou've already chosen the best response.1the 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 \[\large b_nx^n y^{(n)}+\cdots +b_1xy'+b_0y=r(x)\]

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.1you 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
 4 years ago
Best ResponseYou've already chosen the best response.1@RolyPoly hope i have given you peace of mind in this problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My... 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
 4 years ago
Best ResponseYou've already chosen the best response.1that may be the cause of your problems.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Most 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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