Answer:
Explanation:
To solve the differential equation:
y′′′ + y = 3(y′ − 1)
we first find the characteristic equation:
r^3 + 1 = 0
The roots of this equation are:
r = -1, 0 ± i
So, the general solution of the homogeneous equation (y′′′ + y = 0) is:
y_h(t) = c1 e^(-t) + c2 cos(t) + c3 sin(t)
To find a particular solution of the non-homogeneous equation (3(y′ − 1)), we assume a solution of the form:
y_p(t) = a t + b
Taking the first, second and third derivatives of y_p(t), we get:
y′_p(t) = a
y′′_p(t) = 0
y′′′_p(t) = 0
Substituting these into the differential equation, we get:
0 + (a t + b) = 3(a - 1)
Solving for a and b, we get:
a = 1
b = 2
Therefore, the particular solution of the non-homogeneous equation is:
y_p(t) = t + 2
So, the general solution of the non-homogeneous equation is:
y(t) = y_h(t) + y_p(t) = c1 e^(-t) + c2 cos(t) + c3 sin(t) + t + 2
To find the values of c1, c2 and c3, we use the initial conditions:
y(0) = c1 + c2 + 2 = 1
y′(0) = -c1 + c2 + c3 = 0
y′′(0) = c1 - c2 = 0
Solving these equations simultaneously, we get:
c1 = c2 = 1/2
c3 = -1/2
So, the solution of the differential equation with initial conditions (0) = 1, ′(0) = 0, ′′(0) = 0 is:
y(t) = 1/2 e^(-t) + 1/2 cos(t) - 1/2 sin(t) + t + 2