Inconsistency for solving $x' = x^{1/2}$

Question

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.

Answer 1

The solution of $$x'=\sqrt x$$ is given by $$x=\frac{1}{4} \left(t+c\right)^2\implies x'=\color{red}{2 \times}\frac{1}{4} \left(t+c\right) =\frac{1}{2} \left(t+c\right)=\sqrt x$$

Answer 2

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' = $\sqrt{x}$

Answer 3

By separation of variables: $$ \frac{dx}{\sqrt{x}}=dt $$ so $$ \sqrt{x}=\frac{1}{2}(t+c) $$ and finally $$ x=\frac{1}{4}(t+c)^2=\frac{1}{4}(t^2+2ct+c^2) $$ Can you spot your error?

$\displaystyle x=\frac{1}{4}(t^2+{\Large\color{red}{2}}ct+c^2)$