Finding linearly independent solutions

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.
 
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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?
 
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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.
 
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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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