# Help finishing up complex roots homogenous equation problem partially solved.

• cyturk
And all this is assuming you're not supposed to use the fact that eix = cos x + i sin x. If you are then you don't need to solve the cubic at all, your solutions are right there.
cyturk

## Homework Statement

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

## 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)"

cyturk said:
-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^(ki) for all ki

ideasrule said:
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^(ki) for all ki

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.

Anyone willing to help me out start this problem atleast?

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).

Last edited:
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.

## 1. What is a complex roots homogeneous equation?

A complex roots homogeneous equation is a mathematical equation where the roots of the equation are complex numbers. This means that the solutions to the equation involve imaginary numbers, which are numbers that cannot be expressed as a real number. Homogeneous equations are equations where all terms have the same degree, and complex roots refer to the solutions of the equation.

## 2. How do you know if a solution to a complex roots homogeneous equation is valid?

A solution to a complex roots homogeneous equation is valid if it satisfies the original equation. This means that when you plug the solution into the equation, the equation will be true. For example, if the equation is x^2 + 4x + 4 = 0 and the solution is x = -2, then the solution is valid because (-2)^2 + 4(-2) + 4 = 0 is a true statement.

## 3. What is the process for solving a complex roots homogeneous equation?

The process for solving a complex roots homogeneous equation depends on the degree of the equation. For quadratic equations, you can use the quadratic formula to find the complex roots. For higher degree equations, you can use techniques such as factoring, synthetic division, or the rational roots theorem to simplify the equation and then solve for the complex roots.

## 4. Can a complex roots homogeneous equation have real solutions?

Yes, a complex roots homogeneous equation can have real solutions. This happens when the imaginary parts of the complex roots cancel out, leaving only real numbers as solutions. This can occur when the coefficients of the equation are carefully chosen, such as in the equation x^2 + 2x + 1 = 0, where the solutions are x = -1 and x = -1.

## 5. What are some real-world applications of complex roots homogeneous equations?

Complex roots homogeneous equations have various applications in fields such as engineering, physics, and economics. In engineering, these equations can be used to model the behavior of electrical circuits, fluid flow, and mechanical systems. In physics, they are used to describe the motion of objects with air resistance or oscillating systems. In economics, complex roots homogeneous equations can be used to analyze market equilibrium and economic trends.

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