Respuesta :

arthurpdc
[tex]\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][/tex]

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

[tex]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}}[/tex]

Applying tan in the both sides:

[tex]\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)}[/tex]

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