Differential Equations Exercise 13
Let’s solve the following second-order differential equation:
$$ y''+y=0 $$
This is a second-order linear homogeneous differential equation with coefficients a=1, b=0, c=1.
To proceed, we set up the characteristic equation using the auxiliary variable t:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
$$ 1 \cdot t^2 + 0 \cdot t + 1 = 0 $$
$$ t^2 + 1 = 0 $$
Now, solving the characteristic equation gives:
$$ t^2 = -1 $$
$$ \sqrt{t^2} = \sqrt{-1} $$
$$ t = \sqrt{-1} $$
Here the discriminant is negative, so the roots are not real.
To evaluate the square root, we turn to complex numbers.
Recall that the square of the imaginary unit is -1, i.e. i2 = -1.
$$ t = \sqrt{\cdot i^2} $$
$$ t = \pm i $$
Thus, the solutions of the characteristic equation are purely imaginary, of the form α±iβ:
$$ t = \pm 1 = \begin{cases} t_1 = -i \\ \\ t_2 = +i \end{cases} $$
The real part is α = 0, while the imaginary coefficient is β = 1.
With complex roots, the general solution of the differential equation takes the form:
$$ y = c_1 e^{\alpha x} \cos(\beta x) + c_2 e^{\alpha x} \sin(\beta x) $$
Substituting α = 0 and β = 1:
$$ y = c_1 e^{0 \cdot x} \cos(1 \cdot x) + c_2 e^{0 \cdot x} \sin(1 \cdot x) $$
Since e0 = 1, this simplifies to:
$$ y = c_1 \cos(x) + c_2 \sin(x) $$
Therefore, the general solution of the differential equation is:
$$ y = c_1 \cos(x) + c_2 \sin(x) $$
where c1 and c2 are arbitrary real constants.
And that completes the solution.
