Nonlinear Differential Equation for physics I was trying to calculate an equation of motion and there was a differential equation I couldn't solve:
$$
\frac{d^2\theta}{dt^2}+A\cos(\theta)=0 
$$
Does someone know how to solve it?
 A: First, I would substitute $\theta=\pi/2-y$ making the equation 
$$-\frac{d^2y}{dt^2}+A \sin(y)=0 .$$
Secondly I would scale time according to the rule $t=x/\sqrt{A}$ (here I will assume that $A>0$). This will result in the simpler equation
$$\frac{d^2y}{dx^2}=\sin(y) $$
The general solution to this equation, is similar (note the minus sign) to the one that appears in a previous question:
$$y(x)=2\text{am}\left(c_1x +c_2\bigg|-\frac{1}{c_1^2}\right) $$
where am is the Jacobi amplitude function, and $c_1,c_2$ are arbitrary constants. Undoing the substitutions, we obtain the solution of the original equation as $$\theta(t)=\frac{\pi}{2}-2\text{am}\left(c_1\sqrt{A} t +c_2\bigg|-\frac{1}{c_1^2}\right) .$$
A: This is like duffing's equation ,for little value of $\theta$ 
$\frac{d^2\theta}{dt^2}+A\sin (\theta)=0$
to find solution :put $sin (\theta)=\theta-\dfrac{\theta^3}{3!} $ 
and equation becomes $\frac{d^2\theta}{dt^2}+A(\theta-\dfrac{\theta^3}{3!})=0 $ so , you can turn $$\frac{d^2\theta}{dt^2}+A\sin (\theta)=0 \\into \\\frac{d^2\theta}{dt^2}+A(1-\dfrac{\theta^2}{2!})=0$$ and solve numerically
