Differential equation y'""+5y""-2y'"-10y"+y'+5y=0
The teacher will be very surprised to see your correct solution 😉
Solve (d^5 y(x))/(dx^5) + 5 (d^4 y(x))/(dx^4) - 2 (d^3 y(x))/(dx^3) - 10 (d^2 y(x))/(dx^2) + (dy(x))/(dx) + 5 y(x) = 0:
Assume a solution will be proportional to e^(λ x) for some constant λ.
Substitute y(x) = e^(λ x) into the differential equation:
(d^5 )/(dx^5) (e^(λ x)) + 5 (d^4 )/(dx^4) (e^(λ x)) - 2 (d^3 )/(dx^3) (e^(λ x)) - 10 (d^2 )/(dx^2) (e^(λ x)) + d/(dx) (e^(λ x)) + 5 e^(λ x) = 0
Substitute (d^5 )/(dx^5) (e^(λ x)) = λ^5 e^(λ x), (d^4 )/(dx^4) (e^(λ x)) = λ^4 e^(λ x), (d^3 )/(dx^3) (e^(λ x)) = λ^3 e^(λ x), (d^2 )/(dx^2) (e^(λ x)) = λ^2 e^(λ x), and d/(dx) (e^(λ x)) = λ e^(λ x):
λ^5 e^(λ x) + 5 λ^4 e^(λ x) - 2 λ^3 e^(λ x) - 10 λ^2 e^(λ x) + λ e^(λ x) + 5 e^(λ x) = 0
Factor out e^(λ x):
(λ^5 + 5 λ^4 - 2 λ^3 - 10 λ^2 + λ + 5) e^(λ x) = 0
Since e^(λ x) !=0 for any finite λ, the zeros must come from the polynomial:
λ^5 + 5 λ^4 - 2 λ^3 - 10 λ^2 + λ + 5 = 0
Factor:
(λ - 1)^2 (λ + 1)^2 (λ + 5) = 0
Solve for λ:
λ = -5 or λ = -1 or λ = -1 or λ = 1 or λ = 1
The root λ = -5 gives y_1(x) = c_1 e^(-5 x) as a solution, where c_1 is an arbitrary constant.
The multiplicity of the root λ = -1 is 2 which gives y_2(x) = c_2 e^(-x), y_3(x) = c_3 e^(-x) x as solutions, where c_2 and c_3 are arbitrary constants.
The multiplicity of the root λ = 1 is 2 which gives y_4(x) = c_4 e^x, y_5(x) = c_5 e^x x as solutions, where c_4 and c_5 are arbitrary constants.
The general solution is the sum of the above solutions:
Answer: |
| y(x) = y_1(x) + y_2(x) + y_3(x) + y_4(x) + y_5(x) = c_1 e^(-5 x) + c_2 e^(-x) + c_3 e^(-x) x + c_4 e^x + c_5 e^x x