Solving a DE using an ansatz

Finding the first and second derivative of out ansatz, $$\dot x(t)=A(cos(\omega_0 t) - t\omega_0 sin(\omega_0 t)) + B(sin(\omega_0 t) + t\omega_0 cos(\omega_0 t))$$ $$\ddot x= A(-2\omega_0 sin(\omega_0 t) - t{\omega_0}^2cos(\omega_0 t)) + B(2\omega_0 cos(\omega_0 cos(\omega_0 t) -t{\omega_0}^2sin(\omega_0 t)))$$
The differential Equation we are trying to find a solution to is, $$\ddot x + {\omega_0}^2 x = cos(\omega_0 t)$$
if we plug in ##\dot x## and ##\ddot x## and after a little simplification we end up with, $$2\omega_0(Bcos(\omega_0 t) - Asin(\omega_0 t)) = cos(\omega_0t)$$
From here we essentially guess A and B such that the LHS=RHS, I can't think of any possible values that could satisfy the equation.

I didn't check every step in the algebra, but I see A and B that satisfy the last equation.

Hill said:

I didn't check every step in the algebra, but I see A and B that satisfy the last equation.

I initially thought of $$A=sin(\omega_0 t)$$ and $$B=cos(\omega_0 t)$$ using the trig identity, ##cos(2\theta) = cos^2(\theta)-sin^2(\theta)## we get,
$$2\omega_0 cos(2\omega_0 t) = cos(\omega_0 t)$$
but even this doesn't work.

Hamiltonian said:

I initially thought of $$A=sin(\omega_0 t)$$ and $$B=cos(\omega_0 t)$$ using the trig identity, ##cos(2\theta) = cos^2(\theta)-sin^2(\theta)## we get,
$$2\omega_0 cos(2\omega_0 t) = cos(\omega_0 t)$$
but even this doesn't work.

A and B are constants!

Hamiltonian said:

##2\omega_0(Bcos(\omega_0 t) - Asin(\omega_0 t)) = cos(\omega_0t)##

The above has to be true for all values of the variable t, so it should be clear that the absence of a ##\sin(\omega_0 t)## term on the right side has an effect on that term on the left side.

Mark44 said:

The above has to be true for all values of the variable t, so it should be clear that the absence of a ##\sin(\omega_0 t)## term on the right side has an effect on that term on the left side.

Are you implying, ##A=0## and ##B=\frac{1}{2\omega_0}##

Edit: I shall not show my face around these parts of town henceforth.