Help finishing up complex roots homogenous equation problem partially solved.

[cyturk]

Homework Statement

Find the general solution to
y'''+y=0

Homework Equations

The Attempt at a Solution

y''+y=0
r^3+1=0
r^3=-1
r=(-1)^(1/3)
---------------------
-1=-1+0i
-1=cos(pi)+isin(pi)=e^(i*pi)
-1=cos(pi+2xpi)+isin(pi+2xpi)=e^i(pi+2xpi)
--------------------------
(-1)^(1/3)=(e^i(pi+2xpi))^(1/3)
(-1)^(1/3)=(e^i(pi/3+2xpi/3))
(-1)=cos(pi/3+2xpi/3)+isin(pi/3+2xpi/3)
-----------------------------
How do I get this to the general solution form? I know I do something where I let x=0,1,2,3,etc.
But I am not on sure what steps to take.

The answer on Wolfram Alpha is
"y(x) = c_1 e^(-x)+c_2 e^(x/2) sin((sqrt(3) x)/2)+c_3 e^(x/2) cos((sqrt(3) x)/2)"

Thanks in advance!

 

[ideasrule]

-1=-1+0i
-1=cos(pi)+isin(pi)=e^(i*pi)
-1=cos(pi+2xpi)+isin(pi+2xpi)=e^i(pi+2xpi)
--------------------------
(-1)^(1/3)=(e^i(pi+2xpi))^(1/3)
(-1)^(1/3)=(e^i(pi/3+2xpi/3))
(-1)=cos(pi/3+2xpi/3)+isin(pi/3+2xpi/3)

I'm not sure why you're doing these calculations. r^3+1=0 has 3 solutions, not just one. Let's call the solutions k1, k2, and k3. The general solution to the differential equation would be y=c1*e^(k1*x)+c2*e^(k2*x)+c3*e^(k3*x). More broadly, the general solution to any linear homogeneous differential equation is a linear combination of e^(k_i) for all k_i

 

[cyturk]

I'm not sure why you're doing these calculations. r^3+1=0 has 3 solutions, not just one. Let's call the solutions k1, k2, and k3. The general solution to the differential equation would be y=c1*e^(k1*x)+c2*e^(k2*x)+c3*e^(k3*x). More broadly, the general solution to any linear homogeneous differential equation is a linear combination of e^(k_i) for all k_i

I will be honest and say I was following a solution from a book problem, this is a nonbook problem assigned and I am really stuck on it. So I may have all my steps incorrect.

Okay, if r^3+1=0 does have three solutions, k1,k2,k3, is it possible to solve them? Because I feel like that would be the difficult part and I don't know how to do it.

 

[cyturk]

Anyone willing to help me out start this problem atleast?

 

[Dick]

e^(-pi*i), e^(pi*i), e^(3*pi*i), e^(5*pi*i), e^(7*pi*i), e^(9*pi*i), ... are all equal to (-1), right? They are all forms of e^(pi*i+2*pi*i*k) for k an integer. If you take the cube root of all of them by dividing the argument by 3, how many of those numbers are different? You should find that three of them are. Then as ideasrule said those are your k1, k2 and k3. And the general solution is c1*e^(k1*x)+c2*e^(k2*x)+c2*e^(k3*x).

 

[Bohrok]

You could solve r³ + 1 = 0 by factoring the left side first since it's a sum of cubes, using a³ + b³ = (a + b)(a² - ab + b²). Or even easier, you know -1 is a root, so use long or synthetic division to divide r³ + 1 by it and get your quadratic, then solve that to get the last two solutions.