Solving 5th Degree Characteristic Eq. of Linear Homog. Diff. Eq.

In summary, the conversation discusses finding the solution for the nth degree characteristic equation of an nth order linear homogenous differential equation. The conversation also mentions the Abel-Ruffini theorem, which states that there is no general formula for solving fifth or higher degree polynomial equations. Various methods for finding the roots of a polynomial equation are suggested, including the rational root theorem and factoring. Ultimately, it is concluded that solving higher order differential equations can be time-consuming and may require numerical methods.
  • #1
bobmerhebi
38
0

Homework Statement



Given an nth order linear homog. diff eq.

how can I find the solution for its nth degree characteristic eq?

I know its simple Algebra but please help. if possible please give a 5th deg eq. thx
 
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  • #3


The characteristic equation is a 5th degree polynomial equation. As the link Vid provided says, there is NO general formula for solving fifth or higher degree polymnomial equations. You can try, for example, the "rational root theorem" which says that if m/n is a rational number satifying [itex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_0= [/itex], with all coefficients integer then m must evenly divide [itex]a_n[/itex] and n must evenly divide [itex]a_0[/itex]. IF there is a rational root, that at least reduce the possiblilities.
 
  • #4


the thing is if I am given an ordinary linear homog. differe eq of order greater than 2. how should i solve it?

how could i get the roots of the eq. LaTeX Code: a_nx^n+ a_{n-1}x^{n-1}+ \\cdot\\cdot\\cdot+ a_0= 0
in order to find the general sol. of the D.E.

forexample: 4y''' - 3y' + y = 0. this eq. gives 4m3 - 3m + = 0 as a characteristic eq.. how can this low deg. eq be solved? & how can it be applied to higher degree ones. (in other words i need a fast & easy way to find the roots to use in sovlving D.E.'s).

I get that the roots of the above eq. are -1 & 1/2; where 1/2 is a repeated root.
i got the roots by looking @ the divisors of the 1st & last coefficients & dividing the divisors of the last coefficient by the divisors of the 1st one (like +or - 1/4, +/- 1/2, ...) & them checking which makes the polynomial zero. then simplifying the polynomial by dividing it by (x - root found) & finding a low deg polynomial.

but this way takes time, specially if i have a higher order D.E.

so what do u have to say?
 
  • #5


Do what you did. Factor them. If you can't factor them, you are in big trouble as far as solving them exactly. You can always solve them numerically. I'm sorry if it takes time. But that's life.
 

1. What is a 5th degree characteristic equation of a linear homogeneous differential equation?

A 5th degree characteristic equation of a linear homogeneous differential equation is an equation that is formed by setting the coefficients of the highest order derivatives to zero. This equation helps us find the roots, or solutions, of the differential equation.

2. How do you solve a 5th degree characteristic equation of a linear homogeneous differential equation?

To solve a 5th degree characteristic equation, we can use the method of undetermined coefficients or the method of variation of parameters. Both methods involve finding the roots of the equation and using them to construct the general solution of the differential equation.

3. Can all 5th degree characteristic equations be solved analytically?

No, not all 5th degree characteristic equations can be solved analytically. Some equations may have complex roots, making it difficult to find an analytical solution. In these cases, numerical methods may be used to approximate the solution.

4. What is the importance of solving 5th degree characteristic equations of linear homogeneous differential equations?

Solving 5th degree characteristic equations allows us to find the general solution of a linear homogeneous differential equation. This general solution can then be used to solve specific initial value problems, making it an important tool in many areas of science and engineering.

5. Are there any applications of 5th degree characteristic equations in real-world problems?

Yes, 5th degree characteristic equations have many applications in real-world problems, such as in physics, engineering, and economics. For example, they can be used to model the motion of a pendulum, the growth of a population, or the flow of electricity in a circuit.

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