# Implications from Characteristic Equation from Second Order Diffeq

Lets say I have:

$\ y''(t) + 3y'(t) + 4y(t)$ = 0

with characteristic equation

$\ x^2 + 3x + 4$ = 0

What implications about the plot of the solution to the diffeq can be drawn from the characteristic equation? I know that the coefficient to $\ y'(t)$ is the damping term and we can determine if it is over/critically damped by comparing it to a function of the coefficient to the $\ y(t)$ term (don't remember it off the top of my head). What else am I missing?

UPDATE: I know how to solve the diffeq. My question is what implications about the solution can be made from the characteristic equation BEFORE solving all the way. I mean in terms of damping, convergence, resonance, etc.

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Hard to give an explicit answer - but you can write your system easily as $\cdot{y}=...,\cdot{x} = y$ and then discuss its eigenvalues (equation for such eigenvalues will be the same as yours). – Ilya Oct 10 '11 at 16:08

Solutions of the characteristic equation are $x_{1,2} =\frac{-3}{2}\pm\frac{\sqrt 7}{2}i$ so we may conclude that two corresponding solutions of the differential equation are :

$$y_1(t)=e^{(\frac{-3}{2}+\frac{\sqrt 7}{2}i)t} , y_2(t)=e^{(\frac{-3}{2}-\frac{\sqrt 7}{2}i)t}$$

if we apply Euler's formula: $e^{p\pm qi}=e^p(\cos(\pm q)+i\sin(\pm q))$ ,and if we make combination of $y_1(t)$ and $y_2(t)$ we may write following equation which is general solution of the ODE:

$y(t)=e^{pt}(C_1\cos(qt)+C_2\sin(qt))$ where $p=\frac{-3}{2} , q=\frac{\sqrt 7}{2}$

Observing this form you can easily make conclusion about the plot of the solution.

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updated above.. – riotburn Oct 10 '11 at 19:03

Suppose you have $ay''+by'+cy=0$ (some constants $a,b,c$).

Then your characteristic equation is $ax^2+bx+c=0$ and so its roots are $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Looking at real solutions, we have 3 different cases:

1) $b^2-4ac>0$ $\Rightarrow$ distinct real roots so the answer looks like $y=C_1e^{r_1t}+C_2e^{r_2t}$ (over damped)

2) $b^2-4ac=0$ $\Rightarrow$ repeated real root so that answer looks like $y=C_1e^{rt}+C_2te^{rt}$ (critically damped)

3) $b^2-4ac<0$ $\Rightarrow$ conjugate pair of complex roots so your answer looks like $y=C_1e^{\alpha t}\cos(\beta t)+C_2e^{\alpha t}\sin(\beta t)$ (under damped). If $\alpha=-b/2a>0$ then the answer's amplitude grows and grows. If $\alpha=-b/(2a)<0$ then the amplitude gets smaller and smaller.

(See http://en.wikipedia.org/wiki/Harmonic_oscillator Damped harmonic oscillator for more)

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