ODE : not sure how to make it all equal to zero in part (a)? I have found the first and second derivative since we required to do so

First derivative:

Second derivative:

The prof says to substitute the value or r and w at the end but I still cant make it zero
 A: When substituting into the ODE, you should get
$a x''(t) + b x'(t) + c x(t) \\= 
a \left(r^2 e^{r t} c_1 \cos (\omega t)+c_2 \sin (\omega t))+e^{r t} \left(-c_1 w^2 \cos (\omega t)-c_2 \omega^2 \sin (\omega t)\right)+2 r e^{r t} (c_2 \omega \cos (\omega t)-c_1 \omega \sin (\omega t))\right)+b \left(r e^{r t} (c_1 \cos ( \omega t)+c_2 \sin (\omega t))+e^{r t} (c_2 \omega \cos (\omega t)-c_1 \omega \sin (\omega t))\right)+c e^{r t} (c_1 \cos (\omega t)+c_2 \sin (\omega t)) \\=e^{r t} \left(\sin (\omega t) \left(-2 a c_1 r \omega+a c_2 r^2-a c_2 \omega^2-b c_1 \omega+b c_2 r+c c_2\right)+\cos (\omega t) \left(a c_1 r^2-a c_1 \omega^2+2 a c_2 r \omega+b c_1 r+b c_2 \omega+c c_1\right)\right)$
Can you get to this point?
Now, substitute
$$r = -\dfrac{b}{2a}, \omega = \dfrac{\sqrt{4 a c - b^2}}{2a} $$
After the substitution, we get
$a x''(t) + b x'(t) + c x(t) \\= e^{-\dfrac{b t}{2 a}} \left(\left(-\dfrac{c_1 \left(4 a c-b^2\right)}{4 a}-\dfrac{b^2 c_1}{4 a}+c c_1\right) \cos \left(\dfrac{ \sqrt{4 a c-b^2}t}{2 a}\right) \\ +\left(-\dfrac{c_2 \left(4 a c-b^2\right)}{4 a}-\dfrac{b^2 c_2}{4 a}+c c_2\right) \sin \left(\dfrac{\sqrt{4 a c-b^2}t}{2 a}\right)\right)$
Do you now see how each of the terms cancels to zero (just work the cosine by itself and then the sine by itself.
The final result is zero (actually verified it using that expression).
