Second-Order Differential Equation Exercise
Consider the differential equation:
$$ y''+2y'+3y=0 $$
This is a second-order linear homogeneous differential equation with constant coefficients, where a=1, b=2, c=3.
To solve it, we form the characteristic equation using the auxiliary variable t:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
$$ 1 \cdot t^2 + 2 \cdot t + 3 = 0 $$
$$ t^2 + 2t + 3 = 0 $$
Solving this quadratic equation gives:
$$ t = \frac{-2 \pm \sqrt{4-4(1)(3)}}{2} $$
$$ t = \frac{-2 \pm \sqrt{4-12}}{2} $$
$$ t = \frac{-2 \pm \sqrt{-8}}{2} $$
Since the discriminant is negative, we introduce complex numbers to evaluate the square root:
$$ t = \frac{-2 \pm \sqrt{8 \cdot (-1)}}{2} $$
Recalling that the imaginary unit satisfies i2=-1:
$$ t = \frac{-2 \pm \sqrt{8 \cdot i^2}}{2} $$
$$ t = \frac{-2 \pm i \cdot \sqrt{8}}{2} $$
At this point the radicand is positive, so the square root can be simplified:
$$ t = \frac{-2 \pm i \cdot \sqrt{2^3}}{2} $$
$$ t = \frac{-2 \pm i \cdot 2 \sqrt{2}}{2} $$
$$ t = -1 \pm i \sqrt{2} $$
Thus, the characteristic equation has two complex roots:
$$ t = -1 \pm i \sqrt{2} = \begin{cases} t_1 = -1+i\sqrt{2} \\ \\ t_2 = -1-i\sqrt{2} \end{cases} $$
Here the real part is α = -1 and the imaginary coefficient is β = √2.
For complex roots of this form, 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 = -1 and β = \sqrt{2} gives:
$$ y = c_1 e^{-x} \cos(\sqrt{2} x) + c_2 e^{-x} \sin(\sqrt{2} x) $$
This is the general solution of the differential equation, where c1 and c2 are arbitrary real constants.
