Methods to solve $x^2+(x')^2=1$ and similar nonlinear differential equations How do I solve the following:
$x^2 + \left(\frac{dx}{dt}\right)^2 = 1$
Please refer me to an online resource that has worked examples that I can work through to practice.
I have the answer : $ x = \cos{t} $
I need to understand how to get to the answer.
 A: $x^2 + (\frac{dx}{dt})^2 = 1 \Leftrightarrow \frac{dx}{dt} = \sqrt{1-x^2} $ or $\frac{dx}{dt} = -\sqrt{1-x^2} $
Now you have :
$\int \frac{dx}{\sqrt{1-x^2}} = \int dt$ or $ \int -\frac{dx}{\sqrt{1-x^2}} = \int dt $
Which yield you the solutions : $\arcsin(x) = t+c$ or  $-\arcsin(x) = t+c$
Which means : $x(t) = -\sin(t+c_1)$ or $x(t) = \sin(t+c_1)$
Now, depending on your initial conditions, you will be leaded to the answer you mentioned : $x(t) = \cos t$
A: Equations of the type $f(x,x')=0$ or, in general, $f(t,x,x')=0$ are called implicit differential equations. The major technique is to solve the nonlinear algebraic equation first and try to get one or several explicit expressions for $x'$ e.g. of the type $x'=\phi(x)$, resp. $x'=\phi(t,x)$. In some cases, it is convenient to take a parametric solution to $f(x,x')=0$ as $x=g_1(\theta)$, $x'=g_1(\theta)$. For example, to solve your equation one can parameterise $(x,x')$, which is a point on the unit circle, by $\sin\theta$ and $\cos\theta$ as
$$
\begin{cases}
x(t)=\sin\theta(t),\\
x'(t)=\cos\theta(t).
\end{cases}
$$
Then differentiating the first one we get $x'=\theta'(t)\cos\theta(t)$, hence, $\theta'=1$, thus, $\theta(t)=t+c$, or $\cos\theta(t)=0$, thus, $\theta(t)=\pi/2+\pi k$ (the constant solution). 
Note that there is no uniqueness of solution at $x=\pm 1$, so each time the trajectory passes $x=\pm 1$ one can switch to another solution, for example, one can come to $x=1$ along $\sin(t+c)$, then switch to the constant solution $x=1$ for while, and then move along another $\sin(t+c)$. 
You can find some information here and here and perhaps much more to google (like here)
A: $$x^2 + (\frac{dx}{dt})^2 = 1$$
$$ \frac{dx}{dt} = \sqrt{1-x^2}$$
$$\frac{dx}{\sqrt{1-x^2}} = dt$$
$$\int \frac{dx}{\sqrt{1-x^2}} = \int dt$$
$$ \arcsin(x)= t + c $$
$$ x(t) = \sin(t+c)$$
