Euler method with infinite gradient at initial value The title itself is self explanatory - I am trying to numerically solve an ODE with an initial value that has an infinite gradient. It seemed problematic to me and I am not certain as to how I should approach this.
e.g. $\frac {dy}{dx} = \frac y{\sqrt x} , y(0)=1$
(Obviously this can be solved analytically but I would like to know if there is any numerical method that tackles problems like this)
Thank you very much!

Others can probably give a more "conventional" approach but one thing you can do, because $y'(x)$ is independent of $y(x)$, is take the following approach for the first step (at least).
For the initial slope think rather angle. You want to proceed with a certain angle but you can't use $\theta_1=\frac{\pi}{2}$... which is obviously no good.
Instead take as angle the mid-angle of $\theta_1=\pi/2$ and the angle of the curve at $x=h$:
$$\theta_2:=\tan^{-1}\left(y'(h)\right)=\tan^{-1}\left(\frac{1}{\sqrt{h}}\right),$$
and so take as initial angle:
$$\theta_0=\frac{\theta_1+\theta_2}{2}=\frac{\frac{\pi}{2}+\tan^{-1}\left(\frac{1}{\sqrt{h}}\right)}{2},$$
and so an initial slope of:
$$m_0:=\tan(\theta_0).$$
From here you can do normal Euler or perhaps keep it going with slope 
$$m_k=\frac{y'(x_k)+y'(x_{k+1})}{2}.$$
This is an adaptation of Heun's Method. With a step-size of $h=0.1$ this gives an error for $y_1\approx y(x_1)$ of the order of $0.015$.
