How can I find $x(t)$?

Question

given this equation:

$$\frac{d^{2}x}{dt^2} + a \frac{dx}{dt} + x =0$$

and given that: $x(0)=0$, $\frac{dx}{dt}(0)=1 $.

($a$ is parameter and given that $a<2$)

How can I find what is $x(t)$ ? I will be happy for help (or event hints).

Answer 1

You need to solve $$ f''+af+f=0 $$ If you search for exponential function solution, you'll have to solve $$ r^2+ar+1=0 $$ Then $\Delta=a^2-4<0$ because $a<2$. The two solutions are $$ r=\frac{-a \pm i\sqrt{a-2}\sqrt{a+2}}{2} $$ Then the solutions are

$$ f\left(t\right)=e^{-\frac{a}{2}t}\left(\lambda_1\cos\left(\frac{\sqrt{a^2-4}}{2}t\right)+\lambda_2 \sin\left(\frac{\sqrt{a^2-4}}{2}t\right)\right) $$

I let you find the two constants $\lambda_1$ and $\lambda_2$ with initial conditions to find $x$ for all $t$.