Differential Equations Exercise 10
Consider the following second - order differential equation:
$$ y'' - 2y' + y = 0 $$
This is a linear homogeneous differential equation with coefficients a=1, b=-2, and c=1.
The solution begins by forming the characteristic equation, obtained by introducing an auxiliary variable t and substituting the coefficients:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
$$ 1 \cdot t^2 +(-2) \cdot t + 1 = 0 $$
$$ t^2 -2t +1 = 0 $$
Solving this quadratic equation, we find:
$$ t = \frac{-(-2) \pm \sqrt{4-4(1)(1)}}{2} $$
$$ t = \frac{2 \pm \sqrt{4-4}}{2} $$
$$ t = \frac{2 \pm \sqrt{0}}{2} $$
$$ t = \begin{cases} \frac{2+0}{2}=1 \\ \\ \frac{2-0}{2}=1 \end{cases} $$
The characteristic polynomial therefore has a repeated root: t1=1 and t2=1.
In the case of a repeated root, the general solution of the differential equation takes the form:
$$ y = c_1e^{t_1x} + x \cdot c_2e^{t_1x} $$
$$ y = c_1e^{x} + x \cdot c_2e^{x} $$
where c1 and c2 are arbitrary real constants.
This completes the solution.
