Differential Equations Exercise 8
In this exercise, we’re working with a second - order differential equation:
$$ y'' - 5y' + 6y = 0 $$
This is a second - order linear homogeneous differential equation with coefficients a=1, b=-5, and c=6.
To solve it, we first set up the characteristic equation by introducing an auxiliary variable t:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
$$ 1 \cdot t^2 +(-5) \cdot t + 6 = 0 $$
$$ t^2 - 5t +6 = 0 $$
Now we solve this quadratic:
$$ t = \frac{-(-5) \pm \sqrt{25-4(1)(6)}}{2} $$
$$ t = \frac{5 \pm \sqrt{25-24}}{2} $$
$$ t = \frac{5 \pm \sqrt{1}}{2} $$
$$ t = \frac{5 \pm 1}{2} $$
$$ t = \begin{pmatrix} \frac{5+1}{2}=3 \\ \\ \frac{5-1}{2}=2 \end{pmatrix} $$
So the characteristic equation has two roots: t1=3 and t2=2.
Because the roots are distinct, the general solution of the differential equation is:
$$ y = c_1e^{t_1x} + c_2e^{t_2x} $$
$$ y = c_1e^{3x} + c_2e^{2x} $$
where c1 and c2 are arbitrary real constants.
That completes the solution.
