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

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

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    $\begingroup$ 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 $a$. $\endgroup$ – LutzL May 12 at 16:39
  • $\begingroup$ 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. $\endgroup$ – Dylan May 13 at 7:46
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$$ λ^4+2λ^2+a=0 $$ 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.

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