$\displaystyle \frac{d^2x}{dt^2} + \omega^2x = F_0sin(\omega t), x(0) = 0 , x'(0) = 0$

Because this problem is not in term of x and y. Then I assume x is t and y is x.

For general solution,

$\displaystyle x = x_c + x_p$

$\displaystyle m^2 + \omega^2 = 0$

$\displaystyle m = i\sqrt{\omega}$

$\displaystyle x_c = e^t(c_1sin(\sqrt{\omega}t) + c_2cos(\sqrt{\omega}t))$

And for particular solution,

$\displaystyle x_p = Asin(\omega t)$

$\displaystyle x'_p = Acos(\omega t)$

$\displaystyle x''_p = -Asin(\omega t)$

Substitute into $\displaystyle \frac{d^2x}{dt^2} + \omega^2x$ form,

$\displaystyle -Asin(\omega t) + \omega^2Asin(\omega t) = F_0sin(\omega t)$

Then, $\displaystyle A = \frac{F_0}{\omega^2 - 1}$

Finally, $\displaystyle x = e^t(c_1sin(\sqrt{\omega}t) + c_2cos(\sqrt{\omega}t)) + \frac{F_0}{\omega^2 - 1}sin(\omega t)$

My question is how x(0) = 0 and x'(0) = 0 involve with this problem?

Thank you.