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

[GoodEngineering]

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

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

 

[BvU]

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.

 

[Abhishek11235]

$D(D^7-2D^3+1)=0 \implies D=0 or D^7-2D^3+1=0$
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.

 

[BvU]

Can't follow. $D^3 =t \Rightarrow D^7 = t^4 ?$

I would divide by $D-1$, the solution already found.

 

[Ray Vickson]

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.

 

[Mark44]

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 $r = \pm 1, r = \pm i$.

 

[GoodEngineering]

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 Y_c = c₁ + c₂e^x + the remaining products, but as you may know, I need the roots in order to get the whole solution