1. ## Undetermined coefficients

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

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

For general solution,
$x = x_c + x_p$

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

$m = i\sqrt{\omega}$

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

And for particular solution,
$x_p = Asin(\omega t)$

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

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

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

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

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

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

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

2. Originally Posted by noppawit
$\frac{d^2x}{dt^2} + \omega^2x = F_0sin(\omega t), x(0) = 0 , x'(0) = 0$

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

For general solution,
$x = x_c + x_p$

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

$m = i\sqrt{\omega}$

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

And for particular solution,
$x_p = Asin(\omega t)$

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

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

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

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

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

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

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

They are the initial conditions and determine the values of the arbitrary constants $c_1$ and $c_2$ (that is if the general solution was correct which it is not, the solution to the homogeneous equation is wrong ( $m=\pm i \omega$) as is the particular integral)

Spoiler:

$x={\it k_1}\,\sin \left(t\,w\right)+{\it k_2}\,\cos \left(t\,w\right)-{{F_0\,t\,\cos \left(t\,w \right)}\over{2\,w}}$

CB