Finding linearly independent solutions

In summary, the question is asking for three solutions to the differential equation y^(3) + 3y'' + 3y' + y = 0. The problem the person is having is that they do not know how to find these solutions and what they are looking for. They have gone through the process of finding a characteristic equation and found that r^2(r+3) + 3r + 1=0. However, this equation does not have the solutions r=0, r=-3, and r=-1/3. The first equation looks similar to the cube of a binomial, and the second equation is x*e-x. The third equation is y=x*e-x.
  • #1
highlander2k5
10
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 
  • #6
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?
 
  • #7
thanks for the help... I was totally over thinking this question.
 
  • #8
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.
 

1. What does it mean for a set of solutions to be linearly independent?

Linear independence refers to the property of a set of solutions where no one solution can be represented as a linear combination of the others. In other words, each solution in the set adds new information and cannot be duplicated by a combination of the other solutions.

2. Why is it important to find linearly independent solutions?

Finding linearly independent solutions is important because they form the basis for a complete solution space. This means that any solution to the problem can be expressed as a linear combination of these independent solutions. It also simplifies the problem by reducing the number of solutions that need to be considered.

3. How do you determine if a set of solutions is linearly independent?

A set of solutions is linearly independent if and only if the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0 is c1 = c2 = ... = cn = 0, where ci are constants and vi are the solutions. In other words, the only way for the equation to equal zero is if all the constants are zero.

4. Can a set of solutions be both linearly dependent and independent?

No, a set of solutions can only be either linearly dependent or independent. If a set of solutions is linearly dependent, it means that at least one solution can be expressed as a linear combination of the others, making it impossible for the set to be independent.

5. How does the concept of linear independence apply to real-life problems?

In real-life problems, linear independence is used in various fields such as physics, engineering, and economics. It helps to simplify complex problems by reducing the number of solutions that need to be considered. For example, in physics, linearly independent solutions can be used to represent different types of motion or forces acting on a system. In economics, they can be used to represent different variables affecting a market or economy.

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