Deriving a bound for function u(t) in the Cauchy Problem Suppose that $u: [0, τ] \rightarrow \infty$ solves the Cauchy problem: 
$u'(t) = \frac{t^2}{1+u(t)^2}, u(0) = 1$
Find a constant $B ≥ 0$ such that $|u(t)| ≤ B$ for all $t \in [0, τ]$. Your value of B may depend on τ.
So the unit we are learning is error bounds for solutions to the Cauchy problem.
This makes me think a logical first step would be to take the time derivative of $u'(t)$ and try to find an expression for $u''(t)$ where u(t) can be somehow factored out of the resulting expression. 
So I take $u'(t) = f(t,u)$, then use chain rule to find $u''(t)$:
u''(t) = $d \over dt$ u'(t) = $\frac{2t(1+u^2) - u'(t) \cdot 2u \cdot t^2}{(1+u^2)^2}$
Now I plug in $u'(t)$ to simplify the numerator a bit:
$u''(t) = \frac{2t(1+u^2) - \frac{t^2}{1+u^2} \cdot 2u \cdot t^2}{(1+u^2)^2}$ = $\frac{2t(1+u^2)^2 - 2u \cdot t^4}{(1+u^2)^3}$ = ...
At this point I am pretty stuck. I want to extract u = u(t) from this expression, and use information about $t \in [0, τ]$ to bound u(t) but the fraction looks like a mess. Am I missing something or is my approach incorrect? 
For the record, for the same problem except with $f(t, u) = u'(t) = \frac{tu}{1+t^2}$ the exact same approach of differentiating u'(t) works out nicely.
 A: Notice that $u'(t)>0$, thus, $\forall t\ge0\;\; u(t)\ge1$. Integrate both parts of the system:
$$
u(\tau)-u(0) = \int_0^{\tau} \frac{t^2}{1+u(t)^2}\, d\tau
$$
We cannot calculate the integral on the right-hand side, since it contains an unknown function $u(t)$, but we can estimate it:
$$
\int_0^{\tau} \frac{t^2}{1+u(t)^2}\, d\tau\le
\int_0^{\tau} \frac{t^2}{1+1}\, d\tau=\frac{\tau^3}6.
$$
Hence, $u(\tau)\le \frac{\tau^3}6+1$. The function $u(t)$ is strictly increasing (since $u'(t)>0$), thus, $\forall t\in [0, τ]$ we have $1\le u(t)\le \frac{\tau^3}6+1.$
A: Update: I was overthinking the problem and had the wrong idea overall.
Using basic calculus, we can set a bound on u'(t): 
$\lvert u'(t) \rvert = \lvert \frac{t^2}{1+u(t)^2} \rvert ≤ \lvert \frac{t^2}{u(t)^2} \rvert ≤ \lvert \frac{τ^2}{u(t)^2} \rvert = \frac{τ^2}{u(t)^2}$
Since u'(t) ≥ 0 as both the numerator and denominator are non-negative, u(t) is increasing for all $t \in [0, τ]$, meaning that u(0) = 1 is the lowest value u(t) attains. 
Therefore: $\lvert u'(t) \rvert ≤ \frac{τ^2}{u(t)^2} ≤ \frac{τ^2}{u(0)^2} ≤ τ^2$
Now we can bound u(t) with a function f(t) whose derivative is τ^2 and has the same starting point f(0) = 1: $f(t) = τ^2 \cdot t + 1$
So $\lvert u(t) \rvert ≤ \lvert f(t) \rvert = \lvert τ^2 \cdot t + 1 \rvert ≤ τ^3 + 1$ since the max value t attains in [0, τ] is τ
