Solving $x'(t)=\frac{1}{1+x(t)^2} $

Question

I am trying to solve ODE : $x'(t)=\frac{1}{1+x(t)^2} $ with initial value $x(0)=x_0$

I could solve $x'(t)=\frac{1}{x(t)^2}$ by seperating variables.

However, $x'(t)=\frac{1}{1+x(t)^2} $ with initial value $x(0)=x_0$ is another problem, which is I am stucked while studying dynamical system of initial value problem on global system.

Any help would be appreciated.

Answer 1

Hint The equation in the title is still separable: Rewriting gives $$(1 + x^2) \,dx = dt .$$ Integrating gives $$x + \frac{1}{3} x^3 = t + C .$$

Solving the particular i.v.p. with $x(0) = 0$ using, e.g., Cardano's formula gives the particular solution $$x^*(t) = \frac{\left(12 t + 4 \sqrt{4 + 9 t^2}\right)^{1 / 3}}{2} - \frac{2}{\left(12 t + 4 \sqrt{4 + 9 t^2}\right)^{1 / 3}} .$$ Since original system is autonomous, the general solution is $x(t) = x^*(t + C),$ and substituting the initial condition in the last display equation before the spoiler gives $C = x_0 + \frac13 x_0^3$.

Answer 2

The solution being $$\frac 13 x^3+x=t+c_1$$ solve the cubic using the hyperbolic method to obtain as a single root since $$\Delta=-3 (t+c_1)^2-\frac{4}{3} <0$$ and, $$\large\color{blue}{x=2 \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac 3 2 t+c_2 \right)\right)}$$