Method of undetermined coefficients with zero coefficients from initial conditions

[cloud18]

y" + 0.5y' + y = 1-cos(t); y(0) = y'(0) = 0

I used method of undetermined coefficients to get particular solution:

Y(t) = -2sin(t) + 1

To get homogeneous solution, I solved characteristic equation to get complex roots:

r_1,2 = -1/4 +- i*sqrt(15)/4

so homogeneous solution is:

y = c1*exp(-1/4*t)cos(sqrt(15)/4*t) + c2*exp(-1/4*t)sin(sqrt(15)/4*t)

But when I use the initial conditions, I get c1 = c2 = 0.
Is this right, or did I make a mistake?

I simplified the original problem with an easier forcing function to try first, so it is possible the ODE doesn't
make sense?

 

[HallsofIvy]

Yes, you made a mistake! You didn't add the particular solution to the solution to the associated homogeneous equation!
The general solution to the entire equation is y(t)= c1*exp(-1/4*t)cos(sqrt(15)/4*t) + c2*exp(-1/4*t)sin(sqrt(15)/4*t)-2sin(t) + 1

Put x= 0 into that and its derivative.
y(0)= c1+ 1= 0 so c1= -1.

You do the derivative.