When all solutions of $y''+ay'+by=0$ are bounded in R? Could you please help me solve this problem.
Suppose $y''+ay'+by=0$ is differential equation with $a,b$ are real numbers.
I need to find conditions when all solutions of this equation are bounded.
I managed to establish that in case of $a^2-4b \ge 0$ we can found a not bounded solution,
and in case of two pure imaginary roots of characteristic equation all solutions of given equation are bounded.
But I have a difficulty to deal with the case of two complex roots of characteristic equation
$\lambda_1=\alpha+i\beta,\lambda_2=\alpha-i\beta$:
If $\frac {\alpha}{\beta}\ge 1$ we can found a not bounded solution, but what can we do if $\frac {\alpha}{\beta}< 1$ ?
Thanks.
 A: Answer.
If what you are asking is boundedness of ALL the solutions of your ODE, for $t\ge 0$, then the answer is:
$$
a\ge 0\quad\text{and}\quad b>0,
$$
or
$$
a> 0\quad\text{and}\quad b=0,
$$
If what you are asking is boundedness of ALL the solutions, for $t\in \mathbb R$, then the answer is:
$$
a= 0\quad\text{and}\quad b>0.
$$
Explanation. If $b<0$, then the equation $\lambda^2+a\lambda+b=0$ has a positive and negative root, say $r,-s$, and in such case $e^{rt}$ is a non bounded  solution.
If $b=0$, then for $a=0$, the general solution is $c_1t+c_2$, unbounded, while for $a\ne 0$, the general solution is $c_1+c_2e^{-at}$, also unbounded.
If $b>0$, and $a\ne 0$, we have to consider two cases: real roots and complex roots. In both cases we find an unbounded root. 
If $b>0$ and $a=0$, then the general solution is $c_1\cos(\sqrt{b}\,t)+c_2\sin(\sqrt{b}\,t)$, which is of course bounded.
A: The boundedness of the solutions depends on the sign of the real part of the roots of the characteristic equation.
