Initial value problem, $dy/dt=1/y$, $y(0)=0$ i am curious why following initial-value problem
$$ \frac{dy}{dt}=\frac{1}{y},\quad   y(0)=0$$  has no solution
if we solve  it by method of seperation of variables, we get that
$$y(t)=\pm\sqrt{2t\  {}} $$
we  have assumption that our function has form  $f(t,y)$;
book  from which i have taken this example,says that ,it has not solution  because of  it does not contain $t$ variable (or at book  language,does not include $t$ axis)
i need to understand  it  well, as if i met  such type of  problem, i   won't to  mixes  and say  that,it has solution, thanks a lot of,as a additional fact, in book there    is written,if   change  $y(0)=1$, then   $y(t)=\sqrt{2t+1}$, it is defined on this interval
$(-1/2,\infty)$, does it have solution here?if yes  than, $2t$ would be defined  on
 $[0,\infty]$ right? thanks
 A: Let us first solve the problem with initial condition $y(0)=1$. Rewrite our equation in the usual style as $y\,dy=dt$. Integrate. We get $\frac{1}{2}y^2=t+C$, or equivalently $y^2=2t+C'$.  From the condition $y(0)=1$ we obtain $C'=1$. We have arrived at the implicit function $y^2=2t+1$. If we want an explicit expression for $y$, we get the two solutions $y=\sqrt{2t+1}$ and $y=-\sqrt{2t+1}$.
As to where these solutions, implicit or explicit,  are defined, note that there is no problem if $t>-1/2$, since then $2t+1 \gt 0$.  And, (for real solutions) there is a fatal problem if $t<-1/2$. At $t=-1/2$, the derivative of $\sqrt{2t+1}$ is not defined, so technically neither $\sqrt{2t+1}$ nor $-\sqrt{2t+1}$ satisfies the DE at $t=-1/2$. We conclude that there are two solutions, both valid only for $t>-1/2$.
Now let us turn to the initial condition $y(0)=0$. The procedure we used above gives $y^2=2t$. But note that the derivative of $\sqrt{2t}$ is not defined at $t=0$, since there does not exist an open interval about $0$ in which $\sqrt{2t}$ is defined.
We could, by stretching things a little, accept $y=\sqrt{2t}$ and $y=-\sqrt{2t}$ as solutions for $t \gt 0$. We would need to reinterpret the condition $y(0)=0$ as meaning that $\lim_{t\to 0+} y(t)=0$, and to interpret $\frac{dy}{dt}=\frac{1}{y}$ at $t=0$ as meaning that $\lim_{t\to 0+}y\frac{dy}{dt}=1$. That seems to be an interpretation your book does not wish to make.   
A: Before attemting to solve an IVP by analytically or numerically
 of a first-order ODEs in the form
$y\prime= f(x, y),\ \ y(x_0)= y_0$
we should check whether it has a solution or not. If it has a solution, we ask whether it is unique. These questions are answered by the basic existence-uniquness theorem 
which briefly states that
If $f$ and $\frac{\partial f}{\partial y}$
are continuous in a rectangle containing the point $(x_0, y_0)$, then the IVP has a unique solution. There is also a weak condition on $f$ (instead of $\frac{\partial f}{\partial y}$) which is the Lipschitz continuity. 
In your case $(x_0, y_0)=(0, 0)$ and $f(x, y)= 1/y$ where $f$ is even not defined at $y=0$. Examples of such problems including your question you can look at the notes 
http://www.math.ust.hk/~mamu/courses/303/Notes.pdf
