Methods to solve $x^2+(x')^2=1$ and similar nonlinear differential equations

Question

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.

Answer 1

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)

Answer 2

$$x^2 + (\frac{dx}{dt})^2 = 1$$ $$ \frac{dx}{dt} = \sqrt{1-x^2}$$ $$\frac{dx}{\sqrt{1-x^2}} = dt$$ $$\int \frac{dx}{\sqrt{1-x^2}} = \int dt$$ $$ \arcsin(x)= t + c $$ $$ x(t) = \sin(t+c)$$

Answer 3

$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 : $x(t) = \cos t$