Assume that x(t) is a solution: then, as its derivative is non negative, $\forall t$, $x(t) \geq 0$, so $x(1) \geq 0= \tan 0$. Note moreover that $\forall t \geq 1$, $x(t) \geq x'(t)^2 +1$. This forces $x(t)$, for $t \geq 1$ to be biggen than $\tan (t-1)$, which is the solution of the problem $$y(1)=0 \ \ \ y'(t)=y(t)^2+1,$$ and so the solution cannot exist beyond $1+\frac{\pi}2 <3$.

Assume that x(t) is a solution: then, as its derivative is non negative, $\forall t$, $x(t) \geq 0$, so $x(1) \geq 0= \tan 0$. Note moreover that $\forall t \geq 1$, $x(t) \geq x'(t)^2 +1$. This forces $x(t)$, for $t \geq 1$ to be bigger than $\tan (t-1)$, which is the solution of the problem $$y(1)=0 \ \ \ y'(t)=y(t)^2+1,$$ and so the solution cannot exist beyond $1+\frac{\pi}2 <3$.

Assume that x(t) is a solution: then, as its derivative is non negative, $\forall t$, $x(t) \geq 0$, so $x(1) \geq 0= \tan 0$. Note moreover that $\forall t \geq 1$, $x(t) \geq x'(t)^2 +1$. This forces $x(t)$, for $t \geq 1$ to be biggen than $\tan (t-1)$, which is the solution of the problem $$y(1)=0 \ \ \ y'(t)=y(t)^2+1$$, and so the solution cannot exist beyond $1+\frac{\pi}2 <3$.

Assume that x(t) is a solution: then, as its derivative is non negative, $\forall t$, $x(t) \geq 0$, so $x(1) \geq 0= \tan 0$. Note moreover that $\forall t \geq 1$, $x(t) \geq x'(t)^2 +1$. This forces $x(t)$, for $t \geq 1$ to be biggen than $\tan (t-1)$, which is the solution of the problem $$y(1)=0 \ \ \ y'(t)=y(t)^2+1,$$ and so the solution cannot exist beyond $1+\frac{\pi}2 <3$.