y'' + 4y = cos^2(t)

[PinkJazzX]

Here is what I have done so far, using the Method of Undetermined Coefficients:
y''+4y=cos^2(t),y(0)=y' (0)=0
y''+4y=1/2+1/2 cos^2(2t) (power reduction formula)
Y(t)=A+B sin(2t)+C cos(2t)
Y'(t)=2B cos(2t)-2C sin(2t)
Y'' (t)=-4B sin(2t)-4C cos(2t)
[-4B sin(2t)-4C cos(2t)]+4[A+B sin(2t)+C cos(2t)]=1/2+1/2 cos(2t)
4A=1/2+1/2 cos(2t)

Now, it looks like towards the last steps, the B's and C's cancel each other out, leaving only an A term. What should I do from here to find A, B, and C?

 

[tkhunny]

y''+4y=1/2+1/2 cos^2(2t) (power reduction formula)

If we're reducing the power, why is the exponent still in there?

Y(t)=A+B sin(2t)+C cos(2t)

When in doubt, and sometimes when not, OVER define! You need every kind of structure that could contribute to a solution.

\(\displaystyle Y(t) = A + B\cdt\sin(2t) + C\cdot\cos(2t) + D\cdot t\cdot\sin(2t) + E\cdot t\cdot\cos(2t)\)

The facct that everything cancelled out should suggest this as a useful next step.