Differential Equations Exercise 11
We want to solve the following second-order differential equation:
$$ y'' - 2y' + 2y = 0 $$
This is a second-order linear homogeneous differential equation with constant coefficients, where a=1, b=-2, and c=2.
To begin, we set up the characteristic equation using the auxiliary variable t:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
$$ 1 \cdot t^2 + (-2) \cdot t + 2 = 0 $$
$$ t^2 - 2t + 2 = 0 $$
Now, let’s solve this quadratic equation:
$$ t = \frac{-(-2) \pm \sqrt{4 - 4(1)(2)}}{2} $$
$$ t = \frac{2 \pm \sqrt{4 - 8}}{2} $$
$$ t = \frac{2 \pm \sqrt{-4}}{2} $$
Since the discriminant is negative, we need to introduce complex numbers to evaluate the square root.
Recall that -1 is the square of the imaginary unit, i.e. i2 = -1.
$$ t = \frac{2 \pm \sqrt{-1 \cdot 4}}{2} $$
$$ t = \frac{2 \pm \sqrt{i^2 \cdot 4}}{2} $$
$$ t = \frac{2 \pm i \cdot \sqrt{4}}{2} $$
Now the radicand is positive, so we can compute the square root:
$$ t = \frac{2 \pm i \cdot 2}{2} $$
$$ t = \begin{cases} \frac{2+2i}{2} = 1+i \\ \\ \frac{2-2i}{2} = 1-i \end{cases} $$
The characteristic equation therefore has two complex roots: t1 = 1+i and t2 = 1-i.
As a result, the general solution of the differential equation is
$$ y = c_1 e^{\alpha x } \cos (\beta x) + c_2 e^{ \alpha x} \sin (\beta x) $$
Here α=1 and β=1, since the roots can be written as t1 = α + iβ = 1+i and t2 = α - iβ = 1-i.
$$ y = c_1 e^{1 \cdot x } \cos (1 \cdot x) + c_2 e^{ 1 \cdot x} \sin (1 \cdot x) $$
$$ y = c_1 e^{x} \cos (x) + c_2 e^{x} \sin (x) $$
The final expression gives the general solution, where c1 and c2 are arbitrary real constants.
That completes the solution.
