Answer to Question #215842 in Differential Equations for Zerin

Question

solve the problem d^2x/dt^2 +x=8cos2t-4sint x(π/2)=-1 (π/2)=0

Answer 1

Solution

To get general solution of homogeneous equation x₀"+x₀=0 let’s solve characteristic equation r²+1=0.

r_1,2=±i

So x₀(t) = Asin(t)+Bcos(t)

Partial solution x₁(t) may be find in the form x₁(t) = Ccos2t+Dtsint+Etcost. Substituting into equation we’ll get

dx₁/dt = -2Csin2t+Dsint+Dtcost+Ecost-Etsint

d²x₁/dt² = -4Ccos2t-Dcost+Dcost-Dtsint-Esint-Esint-Etcost = -4Ccos2t-Dtsint-2Esint-Etcost

d²x₁/dt² +x₁ = -4Ccos2t-Dtsint-2Esint-Etcost+ Ccos2t+Dtsint+Etcost = -3Ccos2t-2Esint = 8cos2t-4sint

Therefore C = -8/3, D = 0, E = 2 and x₁(t) = -8cos(2t)/3+2tcost

General solution of the given equation is x(t) = x₀(t)+x₁(t) = Asin(t)+Bcos(t)-8cos(2t)/3+2tcost      

x(π/2)=-1 => A+8/3 = -1 => A=-11/3

x’(π/2)=0 => -B-π=0 => B=- π

So x(t) =  -11sin(t)/3 - π cos(t) - 8cos(2t)/3 + 2tcost

Answer

x(t) = -11sin(t)/3 - π cos(t) - 8cos(2t)/3 + 2tcost