Finding linearly independent solutions

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Discussion Overview

The discussion revolves around finding three linearly independent solutions to the differential equation y''' + 3y'' + 3y' + y = 0. Participants explore the characteristic equation and the process of deriving solutions, including the potential for complex roots and the nature of the solutions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about starting the problem and suggests finding a characteristic equation, questioning whether it will involve complex numbers.
  • Another participant clarifies that y^(3) refers to the third derivative, correcting the characteristic equation to r^3 + 3r^2 + 3r + 1 = 0, and challenges the roots proposed by the first participant.
  • A participant later confirms the correct form of the characteristic equation and seeks guidance on breaking it down further.
  • Subsequent posts indicate that the characteristic equation simplifies to (r + 1)^3 = 0, leading to a general solution involving exponential and polynomial terms.
  • Participants inquire about the validity of specific proposed solutions, such as y = e^(-x), y = x*e^(-x), and y = x^2*e^(-x), questioning their linear independence.
  • One participant expresses relief at receiving help, indicating they were overthinking the problem.

Areas of Agreement / Disagreement

There is no clear consensus on the roots of the characteristic equation or the determination of linearly independent solutions. Multiple competing views and uncertainties remain regarding the correct approach and solutions.

Contextual Notes

Participants have not fully resolved the steps to determine linear independence or the implications of the characteristic equation's roots. There are also unresolved questions about the nature of the solutions derived.

highlander2k5
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I'm having trouble understanding what to do for this problem. The question I'm trying to answer is: Find 3 linearly independent solutions to the following differential equation, y^(3) + 3y'' + 3y' + y = 0. I really don't know how to even start this problem and what I'm really looking for. I think I need to try to find a characteristic equation, but I don't know if it will deal with complex numbers or not. So far the only thing I can think of for the characteristic equation is r^2(r+3) + 3r + 1 = 0. Below is what I have so far.

y^(3) + 3y'' + 3y' + y = 0
r^2(r+3) + 3r + 1 = 0
r=0 of degree 2
r=-3 of degree 1
r=-1/3 of degree 1
y(x)=c1 + c2x + c3(e^(-3x)) + c4(e^((-1/3)x))

Can anyone tell me if I'm on the right track and if I've done these steps right so far? Also, what do I need to do next because I'm not sure how I know if I have 3 linearly independent solutions?
 
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Is y^(3) supposed to be y''', the third derivative of y? If so then you're characteristic equation is r3+3r2+3r+1=0. And while this is the same thing as r2(r+3)+3r+1=0 That equation does not have the roots r=0, r=-3, and r=-1/3. Does that first equation look at all familiar maybe the cube of some binomial?
 
yes y^(3) is supposed to be y''' I was told that after y'' your supposed to use y^(n). As for the characteristic function I'm getting
r(r^2 + 3r + 3) + 1 = 0. I'm not sure how to break it down after that.
 
highlander2k5 said:
yes y^(3) is supposed to be y''' I was told that after y'' your supposed to use y^(n). As for the characteristic function I'm getting
r(r^2 + 3r + 3) + 1 = 0. I'm not sure how to break it down after that.

What is (x+1)3?
 
o da, thanks... okay, so I get (r+1)^3 = 0. Now for my general solution I get
y(x) = c1(e^-x) + c2x(e^-x) + c3(x^2)(e^-x). How do I figure out the 3 linearly independent solutions?
 
highlander2k5 said:
o da, thanks... okay, so I get (r+1)^3 = 0. Now for my general solution I get
y(x) = c1(e^-x) + c2x(e^-x) + c3(x^2)(e^-x). How do I figure out the 3 linearly independent solutions?

Is y=e-x a solution?

Is y=x*e-x a solution?

Is y=x2*e-x a solution?

Are the three linearly independent?
 
thanks for the help... I was totally over thinking this question.
 
highlander2k5 said:
thanks for the help... I was totally over thinking this question.

No problem glad to help, I tend to overthink things a lot of the time too.
 

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