# Solve: d^2x//dt^2 + g sin theta t =0 if  theta=g//l , and g and l are constants?

Apr 27, 2018

$x \left(t\right) = - g \sin \left(\frac{g}{1}\right) {t}^{2} / 2 + A t + B$

#### Explanation:

You are asking for the solution to:

$\frac{{d}^{2} x}{{\mathrm{dt}}^{2}} = - g \sin \left(\frac{g}{1}\right)$

$\implies \frac{d x}{\mathrm{dt}} = - g \sin \left(\frac{g}{1}\right) t + A$

$\implies x \left(t\right) = - g \sin \left(\frac{g}{1}\right) {t}^{2} / 2 + A t + B$

Apr 28, 2018

$x = {l}^{2} / g \setminus \sin \theta t + A t + B$

#### Explanation:

We have:

$\frac{{d}^{2} x}{{\mathrm{dt}}^{2}} + g \sin \theta t = 0$ where $\theta = \frac{g}{l}$, a constant

Which we can write as:

$\frac{{d}^{2} x}{{\mathrm{dt}}^{2}} = - g \sin \theta t$

We can "separate the variables" , to get:

$\frac{\mathrm{dx}}{\mathrm{dt}} = \int \setminus - g \sin \theta t \setminus \mathrm{dt}$

And integrating give us:

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{g}{\theta} \cos \theta t + A$

And repeating we get:

$x = \int \setminus \frac{g}{\theta} \cos \theta t + A \setminus \mathrm{dt}$

So that:

$x = \frac{g}{\theta} ^ 2 \sin \theta t + A t + B$

$\setminus \setminus = \frac{g}{\frac{g}{l}} ^ 2 \sin \theta t + A t + B$

$\setminus \setminus = g {l}^{2} / {g}^{2} \setminus \sin \theta t + A t + B$

$\setminus \setminus = {l}^{2} / g \setminus \sin \theta t + A t + B$