Differential Equations - Exercise 6
We want to solve the differential equation:
$$ y'' - y = 0 $$
This is a second-order homogeneous linear differential equation with coefficients a=1, b=0, c=-1.
To proceed, we set up the characteristic equation using the auxiliary variable t:
$$ a \cdot t^2 + b \cdot t + c = 0 $$
Substituting a=1, b=0, c=-1 gives:
$$ t^2 - 1 = 0 $$
Solving for t:
$$ t^2 = 1 $$
$$ t = \pm 1 $$
Thus, the roots of the characteristic equation are t1 = 1 and t2 = -1.
Since these are two distinct real roots, the general solution of the differential equation is:
$$ y = c_1 \cdot e^{t_1 x} + c_2 \cdot e^{t_2 x} $$
$$ y = c_1 \cdot e^{x} + c_2 \cdot e^{-x} $$
This expression represents the general solution of the differential equation,
where c1 and c2 are arbitrary real constants.
And so on.
