y” + ω^2y =f0 sinΏt
The end part has really confused me and I would like some help please.
first solve the homogeneous equation, $\displaystyle y'' + \omega^2 y = 0$ by assuming a solution $\displaystyle y_h = e^{rt}$
then, let $\displaystyle y_p = A \sin \Omega t + B \cos \Omega t$
find $\displaystyle y_p''$ and plug it into the original differential equation and solve for $\displaystyle A$ and $\displaystyle B$ by equating coefficients
your final solution is $\displaystyle y(t) = y_h + y_p$