- #1

Hamiltonian

- 296

- 190

- Homework Statement
- Take the ##ans\ddot atz## ##x(t) = At cos(ω_{0}t) + Btsin(ω_{0}t)## and adjust the constants ##A, B##

to solve the DE bellow in the case ##ω = ω_0##.

$$\ddot x + {\omega_0}^2 x=cos(\omega t)$$

- Relevant Equations
- -

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.

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.