Differential Equations Exercise 9
Consider the following second - order differential equation:
$$ y'' + 4y' + 4y = 0 $$
This is a linear homogeneous differential equation with coefficients a=1, b=4, and c=4.
To solve it, we use the method of the characteristic equation.
The characteristic equation is obtained by introducing an auxiliary variable t and substituting the coefficients a=1, b=4, and c=4:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
$$ 1 \cdot t^2 +4 \cdot t + 4 = 0 $$
$$ t^2 +4t +4 = 0 $$
Solving this quadratic, we find:
$$ t = \frac{-(4) \pm \sqrt{16-4(1)(4)}}{2} $$
$$ t = \frac{-4 \pm \sqrt{16-16}}{2} $$
$$ t = \frac{-4 \pm \sqrt{0}}{2} $$
$$ t = \frac{-4 \pm 0}{2} $$
$$ t = \begin{cases} \frac{-4+0}{2}=-2 \\ \\ \frac{-4-0}{2}=-2 \end{cases} $$
The characteristic polynomial therefore has a repeated root: t1=-2 and t2=-2.
Since the root is repeated, the general solution of the differential equation takes the form:
$$ y = c_1e^{t_1x} + x \cdot c_2e^{t_1x} $$
$$ y = c_1e^{-2x} + x \cdot c_2e^{-2x} $$
where c1 and c2 are arbitrary real constants.
This completes the solution.
