Is there an efficient way of solving $y^{(4)}+2y''+y=0$?

Question

Is there an efficient way of solving $y^{(4)}+2y''+y=0$?

I wanted to use the following theorem:

However, the characteristic equation is unfactorable when its highest factor is 3! So I was thinking about using Laplace transform, by first finding $\hat\Phi(s)=(sI-A)^{-1}$, where $A$ is the companion matrix, and transforming it back into the t-domain. But this is a $4\times 4$ matrix I have to find the inverse to and the row operations get super messy with the $s$ variable involved.

So I am wondering...is there a better way that is around this scope of difficulty? Or do I just fight it out with the algebra and find $\hat\Phi(s)$?

Answer 1

$$y^{(4)}+2y''+y=0$$ The characteristic equation is: $$R^4+2R^2+1=0$$ Substitute $s=R^2$ $$s^2+2s+1=0 \implies (s+1)^2=0$$ Then $$(R^2+1)^2=0 \implies (R^2-i^2)^2=0 \implies (R-i)^2(R+i)^2=0$$ $$\implies R=i, R=-i$$ Mutiplicity 2

And solve the equation. Laplace transform will take you more time here. $$y(x)=c_1\cos x+c_2x\cos x+c_3 \sin x+c_4x \sin x$$

Answer 2

By letting $y=e^{mx}$, in the ODE we get $m=i,i,-i,-i$ (two repeated roots). So $$y=(C_1+c_2x)e^{ix}+(D_1+D_2x)e^{-ix} \implies (E_1+E_2x)\cos x+ (F_1+F_2x)\sin x.$$