Differential Equations Exercise 12
Consider the differential equation:
$$ y''+4y=0 $$
This is a second-order linear differential equation.
Specifically, it is a linear homogeneous equation with coefficients a=1, b=0, c=4.
To solve it, we form the characteristic equation using the auxiliary variable t:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
$$ 1 \cdot t^2 + 0 \cdot t + 4 = 0 $$
$$ t^2 + 4 = 0 $$
Rearranging, we obtain:
$$ t^2 = -4 $$
Taking square roots gives:
$$ t = \sqrt{-4} $$
Since the discriminant is negative, we turn to complex numbers to evaluate the square root:
$$ t = \sqrt{4 \cdot (-1)} $$
Recalling that the imaginary unit satisfies i2=-1:
$$ t = \sqrt{4 \cdot i^2} $$
$$ t = i \sqrt{4} $$
Thus:
$$ t = \pm 2i $$
The characteristic equation therefore has two purely imaginary roots:
$$ t = \pm 2i = \begin{cases} t_1 = -2i \\ \\ t_2 = +2i \end{cases} $$
Here the real part is α=0 and the imaginary part is β=2.
For complex conjugate roots, the general solution of the differential equation is:
$$ y = c_1 e^{\alpha x} \cos(\beta x) + c_2 e^{\alpha x} \sin(\beta x) $$
Substituting α=0 and β=2 yields:
$$ y = c_1 e^{0 \cdot x} \cos(2x) + c_2 e^{0 \cdot x} \sin(2x) $$
Since e0 = 1, this simplifies to:
$$ y = c_1 \cos(2x) + c_2 \sin(2x) $$
Thus, the general solution of the differential equation is:
$$ y = c_1 \cos(2x) + c_2 \sin(2x) $$
where c1 and c2 are arbitrary real constants.
