ODE direction field doesn’t match solution For the ODE $y’=\sqrt y$ and the initial condition $y(0)=0$ the solution is $y=\frac{1}{4}x^2$. This doesn’t match the direction field for the ODE. I wondered why this was happening here? I feel I am missing/overlooking something very basic
 A: On solving the differential equation by variable separable method, we get$$\int\frac{y'}{\sqrt y}=\int dx\implies x-2\sqrt y=C$$The initial condition gives $C=0$. Note that in the solution $x=2\sqrt y$, both $x,y\ge0$. But when you square it to obtain $y=x^2/4$, it becomes defined for negative $x$. The curve over $x<0$ does not actually satisfy the ODE:$$y=x^2/4,x<0\implies \sqrt y=|x/2|=-x/2$$ giving $y'=-\sqrt y$ which is not the original ODE. You are probably seeing the mismatch between direction field and the curve tangents only over $x<0$ (look at the graph).

A: Below I have a plot of the slope field and of $y(x) = \frac14x^2$.



We see that the right half fits the slope field perfectly, but the left half doesn't. So clearly something is wrong with the left half.
Not really. Or, it depends. Note that $\frac14x^2$ perfectly fulfills $(y')^2 = y$. So one could say that the non-fit is a result of choosing the "wrong sign" of the square root to the left of the $x$-intersection.
But also note that $y(x) = 0$ is a perfectly valid solution to your differential equation. And for any $a\in \Bbb R$, setting $y(x) = 0$ for $x\leq a$ and $y(x) = \frac14(x-a)^2$ for $x>a$, is also a valid solution.
So with your initial condition, and with the original differential equation $y' = \sqrt y$, all we can really say is that $a\geq 0$, possibly "$a = \infty$" (i.e. $y(x) = 0$ everywhere). With an initial condition like $y(1) = \frac14$, on the other hand, we fix $a = 0$.
A: This IVP
$$
y'=\sqrt{y}, \quad y(0)=0
$$
does not enjoy uniqueness. (This is because the flux $\sqrt{y}$ is not sufficiently smooth.)
It possesses infinitely many solutions:
$$
y_c(x)=\max\left\{0,\frac{1}{4}(x-c)^2\right\}, \quad c \ge 0.
$$
