Answer to Question #328959 in Differential Equations for Themba

Question

Find the general solution using D-operator

(D + 4)^2x = sihn4t

Answer 1

Solution

For homogeneous equation (D + 4)²x =  0 the characteristic equation is

(λ + 4)²=0 =>  λ_1,2 = -4

So the solution of homogeneous equation is x₀(t) = C₁e^-4t+C₂te^-4t, where C₁, C₂ are arbitrary constants.

For given nonhomogeneous equation

(D + 4)²x = sinh4t  or  (D + 4)²x = 0.5e^4t – 0.5e^-4t     

partial solution may be found in the form

x₁(t)= Ae^4t + Bt²e^-4t        

Substitution this into equation gives

(D + 4)x₁ =8Ae^4t + 2Bte^-4t  =>  

(D + 4)²x₁ = (D + 4)(8Ae^4t + 2Bte^-4t ) = 64Ae^4t + 2Be^-4t  =>         

64Ae^4t = 0.5, 2Be^-4t = -0.5  => A = 1/128, B = -1/4

So general solution of given equation is

x(t) = x₀(t) + x₁(t) = C₁e^-4t+C₂te^-4t + e^4t/128 - t²e^-4t/4