Differential equation with infinite initial value

Question

I know that given an initial value problem \begin{eqnarray} f'(x) &=& G(x,f(x)) \\ f(0) &=& c \end{eqnarray} there is a unique solution for $f$ for any sufficiently nice $G$ (in this case Lipschitz continuous is sufficient). What happens if we have a similar situation but with the initial value $f(0) = \infty$? For example, if $G(x,f(x)) = - f(x)^2$, one finds that the solutions must always have the form $f(x) = \frac1{x+u}$ for some constant $u$, so if we want $f(0) = \infty$, we would take $u=0$. My guess is in general, you could convert an equation \begin{eqnarray} f'(x) &=& G(x,f(x)) \\ f(0) &=& \infty \end{eqnarray} into normal differential equation by substituting $h(x) = 1/f(x)$, which would yield \begin{eqnarray} -\frac{h'(x)}{h(x)^2} &=& G\left(x,\frac1{h(x)}\right) \\ h(0) &=& 0 \end{eqnarray} Is this a good approach? My concern is that it the new function $K(x,h) = -h^2 G(x,\frac1h)$ doesn't seem to have any reason to still be Lipschitz continuous, or indeed continuous at all, when $h = 0$. If we perhaps impose a stricter condition on $G$ is the initial value problem with initial value $\infty$ solvable? In general is there some variant of Peano's existence theorem that could be applied to show existence (but possibly not uniqueness) of a solution?