<s>[INST] Inconsistency for solving $x' = x^{1/2}$ The proposed system $x' = x^{1/2}$ can be solved easily to obtain $x(t) = \frac{1}{4} (t^2 + t c + c^2)$, where $c$ is the integration constant.
However, differentiate the newly-found $x(t)$, one gets: $x(t)' = \frac{1}{2}t+\frac{1}{4}c$. This implies that $x^{1/2} = \frac{1}{2}t+\frac{1}{4}c$. However, 
$$(x^{1/2})^2 = \left(\frac{1}{2}t+\frac{1}{4}c\right)^2 = \frac{1}{4}\left(t^2 + \frac{1}{2}tc + c^2\right) \neq \frac{1}{4} \left(t^2 + t c + c^2\right) = x(t)$$
Does anyone know why the inconsistency occurs? I understand there is another solution, but it is also inconsistent.
 [/INST] You can do it as follows:
$\frac{dx}{dt}$ = $\sqrt{x}$. Rewrite this as follows:
$\frac{dx}{\sqrt{x}}$ = dt.
Let u = $\sqrt{x}$, then du = $\frac{dx}{2\sqrt{x}}$. After this, you'll get 2du = $\frac{dx}{\sqrt{x}}$ = dt, and an indefinite integration gives u = $\frac{1}{2}$(t + c) = $\sqrt{x}$, with c $\in$ $\mathbb{R}$. Finally, x = $\frac{1}{4}$(t+c)^2 = x(t). Take x', and you will see that x' = $\sqrt{x}$
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