Question

Find an implicit and an explicit solution of the given initial-value problem. (use x for x(t).) dx dt = 2(x2 + 1), x(π/4) = 1

Answer 1

$\displaystyle \dfrac{dx}{dt}=2(x^2+1)\\\\ \int_{t_0}^t dt=\int_{x_0}^x\dfrac{dx}{2(x^2+1)}\\\\ t-t_0=\dfrac{1}{2}\int_{x_0}^x\dfrac{dx}{(x^2+1)}\\\\ t-t_0=\dfrac{1}{2}\left[\arctan(x)\right]_{x_0}^x\\\\ t-t_0=\dfrac{1}{2}\left[\arctan(x)-\arctan(x_0)\right]$

We'll use $x(\pi/4)=1$, considering that $x_0=1, t_0=\dfrac{\pi}{4}$:

$t-t_0=\dfrac{1}{2}\left[\arctan(x)-\arctan(x_0)\right]\\\\ t-\dfrac{\pi}{4}=\dfrac{1}{2}\left[\arctan(x)-\arctan(1)\right]\\\\ t-\dfrac{\pi}{4}=\dfrac{1}{2}\left[\arctan(x)-\dfrac{\pi}{4}\right]\\\\ t-\dfrac{\pi}{4}=\dfrac{1}{2}\arctan(x)-\dfrac{\pi}{8}\\\\ t-\dfrac{\pi}{8}=\dfrac{1}{2}\arctan(x)\\\\ \boxed{\arctan(x)=2t-\dfrac{\pi}{4}}$

Applying tan in the both sides:

$\arctan(x)=2t-\dfrac{\pi}{4}\\\\ \tan(\arctan(x))=\tan\left(2t-\dfrac{\pi}{4}\right)\\\\ \boxed{x(t)=\tan\left(2t-\dfrac{\pi}{4}\right)}$