Differential Equations - Exercise 7

Consider the differential equation:

$$ y'' - y' - 6y = 0 $$

This is a second-order equation, since the highest derivative of the unknown function $y$ is $y''$.

It is a second-order homogeneous linear differential equation with coefficients $a = 1$, $b = -1$, and $c = -6$.

The associated characteristic equation, in terms of an auxiliary variable $t$, is:

$$ a t^2 + b t + c = 0 $$

Substituting $a = 1$, $b = -1$, $c = -6$ yields:

$$ t^2 - t - 6 = 0 $$

Note. Here the auxiliary variable is written as $t$, but any symbol (such as $z$ or $u$) would serve the same purpose. It is only a placeholder for solving the quadratic.

Solving the quadratic, we find:

$$ t = \frac{1 \pm \sqrt{1 - 4(1)(-6)}}{2} $$

$$ t = \frac{1 \pm \sqrt{25}}{2} $$

$$ t = \frac{1 \pm 5}{2} $$

$$ t = \begin{cases} \tfrac{1+5}{2}=3 \\[6pt] \tfrac{1-5}{2}=-2 \end{cases} $$

Thus, the characteristic roots are $t_1 = 3$ and $t_2 = -2$.

Since the roots are distinct ($t_1 \ne t_2$), the general solution of the differential equation is:

$$ y = c_1 e^{t_1 x} + c_2 e^{t_2 x} $$

Substituting $t_1 = 3$ and $t_2 = -2$ gives:

$$ y = c_1 e^{3x} + c_2 e^{-2x} $$

Here $c_1$ and $c_2$ are arbitrary real constants.

And so on.