How can I solve the following second order differential equation
$$ \frac{d^2\theta \:}{dt^2\:}\:=-\frac{g\theta \:\:}{L} $$ with initial conditions $$θ(0) = θ_0 , v(0) = v_0 $$
I know that I can rewrite the equation as
$$ \frac{d^2\theta \:}{dt^2\:}\:+\frac{g\theta \:\:}{L} = 0 $$
, find characteristic polynomial and solve the equation. I am just confused on how to solve this problem because of all the constants. Thanks for help in advance
$$\frac{d^2\theta \:}{dt^2\:}\:+\frac{g\theta \:\:}{L} = 0$$ Characteristic polynomial is $$r^2+\frac g L=0 \implies r=\pm i \sqrt {\frac gl}$$ Then the solution is $$\theta(t)=c_1\cos(\sqrt {\frac gl}t)+c_2\sin(\sqrt{\frac gl}t)$$ Apply initial conditions now
$$θ(0) = θ_0 ,\implies \theta(0)=c_1 \implies c_1=\theta_0$$ $$\theta(t)=\theta_0\cos(\sqrt{\frac gl}t)+c_2\sin(\sqrt{\frac gl}t)$$ Differentiate $$\theta(t)'=\sqrt{\frac gl} \left(-\theta_0\sin(\sqrt{\frac gl}t)+c_2\cos(\sqrt{\frac gl}t) \right )$$ Apply the second inditial condition to find $c_2$ $$\theta(0)'=\sqrt{\frac gl} c_2 \implies c_2=v_0\sqrt{\frac lg}$$ Finally $$\theta(t)=\theta_0\cos \left(\sqrt{\frac gl}t \right)+v_0\sqrt{\frac lg}\sin \left (\sqrt{\frac gl}t \right)$$