Solve the ODE. y***+y*=e^(-t) where y(0)= pi-(1/2) , y*(0)=1, y**(0)=(-1/2) The

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y' ' '+y'=e^(-t) where y(0)= pi-(1/2) , y*(0)=1, y**(0)=(-1/2) y' ' '+y'=e^(-t) homogeneous solution is y' ' '+y'= 0 ==> m^3 +m=0 ===> m(m^2 +1) =0 m=0 , m=-i , m=i remember that e^(ix) = cos x + i sinx e^(-ix) = cos x - i sinx yh(t) = A e^( 0 t) + B cos(t) + C sin(t) ==> yh(t) = A + B cos(t) + C sin(t) particular solution yp(t) = 1/ (D (D^2 +1)) ) e^(-t) = - e^(t) / (D^2 +1 -1) = - e^(t) / (D^2) = -t^2 - e^(t) general solution is y(t) = A + B cos(t) + C sin(t) -t^2 - e^(t) (Eqn 1001) PUT initial conditions y(0)= pi-(1/2) , y*(0)=1, y**(0)=(-1/2) in Eqn 1001) AND GET A, B, C

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