4.5k views
5 votes
Find the general solution of the given

higher-order differential equation.

y``` - 4y`` - 5y` = 0

User NewBike
by
7.1k points

1 Answer

4 votes

Answer:

Explanation:

The characteristic equation of the given differential equation is:

r^3 - 4r^2 - 5r = r(r^2 - 4r - 5) = r(r - 5)(r + 1) = 0

Thus, the roots of the characteristic equation are r = 0, r = 5, and r = -1.

Therefore, the general solution of the differential equation is:

y(t) = c1 e^(0t) + c2 e^(5t) + c3 e^(-t)

Simplifying this expression, we get:

y(t) = c1 + c2 e^(5t) + c3 e^(-t)

where c1, c2, and c3 are constants determined by the initial conditions of the problem.

User Olga Botvinnik
by
8.5k points