Question

Find the solution of y" – 4y + 4y = 16e^6t 

with y(0) = 1 and y'(0) = 3. 

y = _______ 

Answer 1

Find the complementary solution by solving ( d^2 y(t))/( dt^2) - 4 ( dy(t))/( dt) + 4 y(t) = 0:
Assume a solution will be proportional to e^(λ t) for some constant λ.
Substitute y(t) = e^(λ t) into the differential equation:
( d^2 )/( dt^2)(e^(λ t)) - 4 d/( dt)(e^(λ t)) + 4 e^(λ t) = 0
Substitute ( d^2 )/( dt^2)(e^(λ t)) = λ^2 e^(λ t) and d/( dt)(e^(λ t)) = λ e^(λ t):
λ^2 e^(λ t) - 4 λ e^(λ t) + 4 e^(λ t) = 0
Factor out e^(λ t):
(λ^2 - 4 λ + 4) e^(λ t) = 0
Since e^(λ t) !=0 for any finite λ, the zeros must come from the polynomial:
λ^2 - 4 λ + 4 = 0
Factor:
(λ - 2)^2 = 0
Solve for λ:
λ = 2 or λ = 2
The multiplicity of the root λ = 2 is 2 which gives y_1(t) = c_1 e^(2 t), y_2(t) = c_2 e^(2 t) t as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(t) = y_1(t) + y_2(t) = c_1 e^(2 t) + c_2 e^(2 t) t
Determine the particular solution to ( d^2 y(t))/( dt^2) - 4 ( dy(t))/( dt) + 4 y(t) = 16 e^(6 t) by the method of undetermined coefficients:
The particular solution to ( d^2 y(t))/( dt^2) - 4 ( dy(t))/( dt) + 4 y(t) = 16 e^(6 t) is of the form:
y_p(t) = a_1 e^(6 t)
Solve for the unknown constant a_1:
Compute ( dy_p(t))/( dt):
( dy_p(t))/( dt) = d/( dt)(a_1 e^(6 t))
= 6 a_1 e^(6 t)
Compute ( d^2 y_p(t))/( dt^2):
( d^2 y_p(t))/( dt^2) = ( d^2 )/( dt^2)(a_1 e^(6 t))
= 36 a_1 e^(6 t)
Substitute the particular solution y_p(t) into the differential equation:
( d^2 y_p(t))/( dt^2) - 4 ( dy_p(t))/( dt) + 4 y_p(t) = 16 e^(6 t)
36 a_1 e^(6 t) - 4 (6 a_1 e^(6 t)) + 4 (a_1 e^(6 t)) = 16 e^(6 t)
Simplify:
16 a_1 e^(6 t) = 16 e^(6 t)
Equate the coefficients of e^(6 t) on both sides of the equation:
16 a_1 = 16
Solve the equation:
a_1 = 1
Substitute a_1 into y_p(t) = a_1 e^(6 t):
y_p(t) = e^(6 t)
The general solution is:
y(t) = y_c(t) + y_p(t) = e^(6 t) + c_1 e^(2 t) + c_2 e^(2 t) t
Solve for the unknown constants using the initial conditions:
Compute ( dy(t))/( dt):
( dy(t))/( dt) = d/( dt)(e^(6 t) + c_1 e^(2 t) + c_2 e^(2 t) t)
= 6 e^(6 t) + 2 c_1 e^(2 t) + c_2 e^(2 t) + 2 c_2 e^(2 t) t
Substitute y(0) = 1 into y(t) = e^(6 t) + e^(2 t) c_1 + e^(2 t) t c_2:
c_1 + 1 = 1
Substitute y'(0) = 3 into ( dy(t))/( dt) = 6 e^(6 t) + 2 e^(2 t) c_1 + e^(2 t) c_2 + 2 e^(2 t) t c_2:
2 c_1 + c_2 + 6 = 3
Solve the system:
c_1 = 0
c_2 = -3
Substitute c_1 = 0 and c_2 = -3 into y(t) = e^(6 t) + e^(2 t) c_1 + e^(2 t) t c_2:
Answer: |
| y(t) = e^(6 t) - 3 e^(2 t) t