$y'=k+\sqrt{y},y(0)=0$ Prove $\forall k\neq0$ there is no solution. $y'=k+\sqrt{y},y(0)=0$
Prove $\forall k\neq0$ there is no solution.
My solution:
$\frac{dy}{dx}=k+\sqrt{y}\implies \frac{dy}{k+\sqrt{y}}=dx \implies \int\frac{dy}{k+\sqrt{y}}=\int dx \implies 2\left(\sqrt{y}-k\ln\left(\left|\sqrt{y}+k\right|\right)\right)=x+C$
The initial value is $(0,0)$ , hence $C=-2k\ln|k|$
The solution is $2\left(\sqrt{y}-k\ln\left(\left|\sqrt{y}+k\right|\right)\right)=x-2k\ln|k|$
I don't get why there is no solution.
Any help is welcome.
Thanks !
 A: When $y= 0$ (as at the point $x=0$), the differential equation says $y' = k \ne 0$.
Since $\sqrt{y}$ doesn't exist (as a real number) when $y < 0$, that means that a (real) solution can only be defined for $x \ge 0$ or $x \le 0$.  Since in order for a function to be differentiable at $x=0$ it must be defined in a neighbourhood of $0$ (including both positive and negative numbers), the differential equation is technically not defined at the initial point $y(0)=0$.
But that is really just a technical quibble: people solve differential equations all the time with initial conditions such as this, and it's interpreted as meaning: $y$ satisfies the differential equation for $x > 0$ and $\lim_{x \to 0+} y(x) = 0$.
A: This answer is temporarily out of service.
So if $k=0$, the IVP has solution $y(x)=\left(\frac{x}{2}\right)^{2}$ for all $x\in \mathbb{R}$. If $k\not=0$, then the ODE $y'(x)=k+\sqrt{y(x)}$ is a Chini's equation and we can solve it for this special case as a separable equation and we get $-2k\ln|k+\sqrt{y(x)}|+2\sqrt{y(x)}=x+C$, with $C$ a constant of integration. Setting $y(0)=0$, we get the particular solution for the IVP as $y(x)=k^{2}\left(W\left(-\sqrt{\frac{1}{k^{2}}}ke^{-\frac{x}{2k}-1}\right)+1\right)^{2}$, where $W(\cdot)$ is Lambert W function.
