Find a particular solution to y′′+8y′+16y=4.5e(−4t)
Auxiliary eqn:
m^2 + 8m + 16 = 0
(m + 4)^2 = 0
m = -4 (twice – repeated root)
Complimentary solution is yc = e^(-4t)(At + B)
Particular solution
Since e^(-4t) is in the rhs of the DE, and we already have e(-4t) and t.e^(-4t) in the complimentary solution, let us assume a solution of the form y = C.t^2.e^(-4t)
Then y’ = C.2t.e^(-4t) -4C.t^2.e^(-4t) = C.e^(-4t)(2t – 4t^2)
And y’’ = -4C.e^(-4t)(2t – 4t^2) + C.e^(-4t)(2 – 8t) = C.e^(-4t)(-8t + 16t^2 + 2– 8t)
y’’ = C.e^(-4t)(16t^2 – 16t + 2)
Substituting the expressions for y, y’ and y’’ into the differential equation,
C.e^(-4t)(16t^2 – 16t + 2)+ 8 C.e^(-4t)(2t – 4t^2) + 16 C.t^2.e^(-4t) = 4.5e^(-4t)
Cancelling out e^(-4t),
C.(16t^2 – 16t + 2) + 8C.(2t – 4t^2) + 16 C.t^2 = 4.5
C.(16t^2 – 16t + 2 + 16t – 32t^2 + 16t) = 4.5
C.(2) = 4.5
C = 2.25
Particular solution is: yp = 2.25.t^2.e^(-4t)
General solution is y = yc + yp
Answer: y = e^(-4t)(At + B) + 2.25.t^2.e^(-4t)