What method should I use to get the roots of this equation?

In summary, the conversation involved a student discussing a calculus test where they had to find the general solution for a differential equation. They had difficulty finding the remaining roots and asked for help. Various methods were suggested, including using solutions of the form y=e^λx and applying Descartes' rule of signs. The conversation also discussed the possibility of a mistake in the original equation given on the test.
  • #1
GoodEngineering
5
1
Mentor note: Thread moved to Diff. Equations subforum
Hello, few days ago I had a calculus test in which I had to find the general solution for the next differential equation: (D^8 - 2D^4 + D)y = 0.

"D" stands for the differential "Dy/Dx". However I could only find 2 of the roots on the ecuation, x1 = 0 and x2 = 1.

By the time I took the test I only knew 3 methods, the P/Q Descartes Method, factorization & the legendary and well known quadratic formula.
Then when I got home I entered the differential equation on MATLAB & got the next roots

241881
So can anyone tell me how to get the remaining roots or what method I have to use?
 
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  • #2
GoodEngineering said:
I only knew 3 methods
To solve differential equations ?

Did it occur to you to try solutions of the form ##y = e^{\lambda x}##, one of the more basic methods for solving differential equations ?

##y(x) = 0 ## and ##y(x) = 1## are indeed solutions, in fact any constant will do. ##x_1=0## is meaningless as a solution of the differential equation.
 
  • #3
##D(D^7-2D^3+1)=0 \implies D=0 or
D^7-2D^3+1=0##
To solve second put ##D^3=t## so that ## t^4-2t+1=0##. Use the Ferrari method to solve this biquadratic equation(Google for the steps). Now what you got is ##D^3##, so, solve it for D by taking complex cube root of the roots you got.
 
  • #4
Can't follow. ##D^3 =t \Rightarrow D^7 = t^4 ?##

I would divide by ##D-1##, the solution already found.
 
  • #5
GoodEngineering said:
Hello, few days ago I had a calculus test in which I had to find the general solution for the next differential equation: (D^8 - 2D^4 + D)y = 0.

"D" stands for the differential "Dy/Dx". However I could only find 2 of the roots on the ecuation, x1 = 0 and x2 = 1.

By the time I took the test I only knew 3 methods, the P/Q Descartes Method, factorization & the legendary and well known quadratic formula.
Then when I got home I entered the differential equation on MATLAB & got the next roots

View attachment 241881So can anyone tell me how to get the remaining roots or what method I have to use?

I doubt that any reasonable method of getting actual solutions is possible on an exam or test. However, you could apply Descartes' rule of signs to determine the number of positive and negative real roots of ##D^7 - 2D^3 + 1##: one negative and two positive real roots ##-r_1, r_2, r_3##. Then there must be four complex roots, coming in conmplex-conjugate pairs: ##s_1 \pm i w_1, s_2 \pm i w_2.## That would allow you to write
$$y' = a_1 e^{-r_1 t} + a_2 e^{r_2 t} + a_3 e^{r_3 t}
+ b_1 \cos(w_1 t) e^{s_1 t} \\
\hspace{4em}
+ b_2 \sin(w_1 t) e^{s_1 t} + c_1 \cos(w_2 t) e^{s_2 t} + c_2 \sin(w_2 t) e^{s_2 t},$$
where the ##a_i, b_i , c_i, r_i>0, s_i, w_i > 0## are real constants. Then ##y = k + \int y' \, dt,## where ##k## is another constant.
 
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  • #6
GoodEngineering said:
Hello, few days ago I had a calculus test in which I had to find the general solution for the next differential equation: (D^8 - 2D^4 + D)y = 0.
Could it be that you are mistaken in what you remember? Asking a student to solve ##y^{(8)} - 2y^{(4)} + y' = 0## on a timed exam is unreasonable, IMO, but asking him or her to solve ##y^{(8)} - 2y^{(4)} + y = 0## is reasonable.

The first equation above has a characteristic equation of ##r^8 - 2r^4 - r = 0##, which can be factored into ##r(r^7 - 2r^3 + 1) = 0##. @Abhishek11235's suggestion in post #3 is no help, as has been pointed out.

The second equation has a characteristic equation of ##r^8 - 2r^4 + 1 = 0##, which can easily be factored into ##(r^4 - 1)^2 = 0## and further to ##(r^2 - 1)^2(r^2 + 1)^2 = 0##, with repeated roots of ##r = \pm 1, r = \pm i##.
 
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  • #7
BvU said:
To solve differential equations ?

Did it occur to you to try solutions of the form ##y = e^{\lambda x}##, one of the more basic methods for solving differential equations ?

##y(x) = 0 ## and ##y(x) = 1## are indeed solutions, in fact any constant will do. ##x_1=0## is meaningless as a solution of the differential equation.
Yes, actually that was the final part on the test, to get the Yc = c1 + c2ex + the remaining products, but as you may know, I need the roots in order to get the whole solution
 

1. What is the best method to use for finding the roots of an equation?

The best method to use for finding the roots of an equation depends on the specific equation and its characteristics. Some commonly used methods include the quadratic formula, factoring, and the Newton-Raphson method. It is important to consider the complexity of the equation and the availability of resources before deciding on a method.

2. How do I know which method is most appropriate for my equation?

To determine the most appropriate method for finding the roots of an equation, it is important to first understand the properties of the equation. This includes the degree of the equation, the number of roots, and any patterns or symmetries. It is also helpful to consult with a mathematics expert or use online resources to compare and contrast different methods.

3. Can I use any method to find the roots of an equation?

While there are many methods available for finding the roots of an equation, not all methods may be suitable for a particular equation. Some methods may work better for certain types of equations or may require certain conditions to be met. It is important to carefully consider the equation and the capabilities of each method before making a decision.

4. Are there any limitations or drawbacks to using certain methods for finding roots?

Yes, some methods may have limitations or drawbacks when used to find the roots of an equation. For example, the quadratic formula may not work for equations with complex roots, while the Newton-Raphson method may require a good initial guess to converge to the correct root. It is important to be aware of these limitations and choose a method that is appropriate for the equation at hand.

5. Is there a "best" method for finding the roots of an equation?

There is no one "best" method for finding the roots of an equation, as different methods may be more suitable depending on the specific equation and its properties. It is important to evaluate the equation and consider factors such as accuracy, efficiency, and ease of use when selecting a method. In some cases, a combination of methods may be the most effective approach.

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