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 (dx)/(dt)=2(x^2+1)\\\\ \int_(t_0)^t dt=\int_(x_0)^x(dx)/(2(x^2+1))\\\\ t-t_0=(1)/(2)\int_(x_0)^x(dx)/((x^2+1))\\\\ t-t_0=(1)/(2)\left[\arctan(x)\right]_(x_0)^x\\\\ t-t_0=(1)/(2)\left[\arctan(x)-\arctan(x_0)\right]`

We'll use
`x(\pi/4)=1`, considering that
`x_0=1, t_0=(\pi)/(4)`:


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

Applying tan in the both sides:


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