Laplace Transform Method:
Taking the Laplace Transform of both sides of the equation, we get:
L{D³y(t)} + 9L{D²y(t)} + 25L{Dy(t)} + 17L{y(t)} = 5L{cos(3t)}
Using the properties of Laplace Transform, we get:
s³Y(s) - s²y(0) - sy'(0) - y''(0) + 9s²Y(s) - 9sy(0) - 9y'(0) + 25sY(s) - 25y(0) + 17Y(s) = 5[s/(s²+9)]
Substituting the initial conditions, we get:
s³Y(s) + 9s²Y(s) + 25sY(s) + 17Y(s) = 5s/(s²+9) + s²
Simplifying, we get:
Y(s) = [5s + s²(s²+9)]/[(s²+9)(s³+9s²+25s+17)]
Using partial fraction decomposition, we get:
Y(s) = -1/(s+3) + 2/(s²+9) + (3s-2)/(s²+4s+13) - (s+1)/(s²+4s+13) - 1/(s-1)
Taking the inverse Laplace Transform, we get the total solution:
y(t) = -e^(-3t) + 2sin(3t) + (3/2)e^(-2t)sin(t) - (1/2)e^(-2t)cos(t) - e^t
Classical Method:
The characteristic equation of the differential equation is:
r³+9r²+25r+17 = 0
Using synthetic division, we get:
(r+1)(r²+8r+17) = 0
The roots are:
r1 = -1, r2 = -4+3i, r3 = -4-3i
Hence, the homogeneous solution is:
y_h(t) = c1e^(-t) + c2e^(-4t)cos(3t) + c3e^(-4t)sin(3t)
To find the particular solution, we assume a solution of the form:
y_p(t) = A cos(3t) + B sin(3t)
Taking the derivatives and substituting in the differential equation, we get:
-8A/3 - 8B = 0
-25A/9 + 25B/3 = 5/9
Solving the system of equations, we get:
A = 5/18, B = -5/54
Hence, the particular solution is:
y_p(t) = (5/18)cos(3t) - (5/54)sin(3t)
The total solution is the sum of the homogeneous and particular solutions:
y(t) = y_h(t) + y_p(t)
Substituting the initial conditions, we get:
c1 = 1, c2 = 0, c3 = 1/3
Hence, the total solution is:
y(t) = e^(-t) + (1/3)e^(-4t)sin(3t) + (5/18)cos(3t) - (5/54)sin(3t)