The given equation is a second-order linear homogeneous differential equation:

$y'' + 4y' + 5y = 0$

To solve this, we'll look for a general solution of the form $y = e^{rt}$, where $r$ is a constant to be determined. Substituting $y = e^{rt}$ into the differential equation:

$r^2 e^{rt} + 4r e^{rt} + 5 e^{rt} = 0$

Factoring out $e^{rt}$ (which is never zero):

$e^{rt} (r^2 + 4r + 5) = 0$

Thus, we solve the characteristic equation:

$r^2 + 4r + 5 = 0$

Using the quadratic formula:

$r = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}$

$r = \frac{-4 \pm 2i}{2} = -2 \pm i$

So, the roots are $r = -2 + i$ and $r = -2 - i$, which are complex conjugates.

The general solution for the differential equation when the roots are complex is:

$y(t) = e^{-2t} (C_1 \cos(t) + C_2 \sin(t))$

Where $C_1$ and $C_2$ are constants determined by initial conditions (if given).

Would you like to go over the solution in more detail, or clarify anything?

Related Questions:

1. How can we solve the equation with initial conditions $y(0) = 1$ and $y'(0) = 0$?
2. What is the interpretation of the complex roots in the context of this differential equation?
3. How can the solution be visualized as a damped oscillation?
4. What would change if the equation had real roots instead of complex ones?
5. How does the behavior of solutions to second-order linear differential equations depend on the discriminant of the characteristic equation?

Tip:

Complex roots indicate oscillatory behavior, often seen in systems involving damping or wave-like motion.