# Behaviour of the solution of ODE for $t \to \infty$

$$y^{(4)}(t)+2y''(t)+ay(t)=0, \ \ \ a \in \mathbb R$$

Determine the general solution of the ODE and its behaviour for $$t \to \infty$$.

The roots of the characteristic polynomial $$\lambda^4+2\lambda^2+a$$ are

$$\lambda_1=-\sqrt{-1-\sqrt{1-a}}\\\lambda_2=\sqrt{-1-\sqrt{1-a}}\\\lambda_3=-\sqrt{-1+\sqrt{1-a}}\\\lambda_4=\sqrt{-1+\sqrt{1-a}}$$

so $$y(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}+c_3e^{\lambda_3t}+c_4e^{\lambda_4t}$$

Now for the behaviour for $$t \to \infty$$: Is there some clever way to do this?

Can I work with the complex fundamental system? Or do I have to look at the different cases for $$a$$ and build a real fundamental system? Or what would be the 'best' way to solve this? It seems like the cases would take too long

• You need to understand that those roots are complex numbers and you need to explore which quadrant they fall into. // Your equation does not contain $a$. – LutzL May 12 at 16:39
• You need to separate cases $a < 1$, $a=1$, $a>1$. For every case, these roots will be complex, and the behavior depends on their real parts. – Dylan May 13 at 7:46

$$λ^4+2λ^2+a=0$$ has for $$a<0$$ one sign change in the coefficient sequence and thus by Descartes rule of signs always a positive root. This means that in general the ODE has unbounded solutions for $$t\to\infty$$.
For $$a>1$$ one gets by completing the square in the first and last term the real factorization $$(λ^2+\sqrt{a})^2-2(\sqrt{a}-1)λ^2 =\left(λ^2+\sqrt{2(\sqrt{a}-1)}λ+\sqrt{a}\right) \left(λ^2-\sqrt{2(\sqrt{a}-1)}λ+\sqrt{a}\right)$$ The second factor will always give at least one root with positive real part, giving again unbounded solutions for the ODE.
For $$a\in[0,1]$$ use standard square completion $$(λ^2+1)^2=1-a\ge 0\implies λ^2=-1\pm\sqrt{1-a}.$$ For $$a\in(0,1)$$ this gives 4 simple imaginary roots so that all solutions of the ODE are bounded. In the case $$a=0$$ or $$a=1$$ the double roots (at $$0$$ resp. $$\pm i$$) will cause basis solutions with linearly growing amplitude to appear.