How to prove that initial value problem have no solution? In my exam I got the following differential equation,
$ty'-y^{-1}= 0$  
where $y'=\frac{dy}{dt}$
And one of the question asked, if a solution exist so that $y(0) = 1$ 
How could I show that for the following differential equation, with the given initial value problem there is no solution? 
 A: Why not just solve it?
$$t\frac{dy}{dt}-y^{-1}=0 \implies y \frac{dy}{dt}=\frac 1t \implies ydy=\frac{dt}{t}\implies \frac {y^2}{2}=\ln At$$
At $t=0$, the solution blows up, hence it is not possible that $y(0)=1$
A: First we solved the differential equation we will get,
$t\frac{dy}{dt}-y^{-1}=0 \Longrightarrow \frac{t}{dt} = \frac{1}{y*dy}$
Now I will make the assumption that $t\neq0$, then I will get that 
$\frac{dt}{t} = y\,dy \Longrightarrow \int{\frac{1}{t}\,dt} = \int{y\,dy} \Longrightarrow ln(t)+c_1=\frac{y^2}{2}+c_2 \Longrightarrow y^2=2ln(t)+2(c_1-c_2)$
And I will set $C = 2(c_1-c_2)$ , and I will get the following general solution of the differential equation,
$y = \pm \sqrt{2ln(t)+C}$ 
To find this solution I made the assumption that $t\neq0$, so in principal there might be other solutions that pass points where $t = 0$. 
Let assume there is such a solution, we know that this solution differentiable so it must be be continues in some neighbourhood of $t = 0$. So this solution, will be also a solution for initial value problem $y(t_0)=y_0$ where $t_0\neq 0$. We know that using the general solution we found earlier we could find a solution for this initial value problem. So we now have "two solutions", one solution that pass through point where $t=0$, and the second when is the one we got from the general solution. From the theory of existence and uniqueness we know that this initial value problem have only one solution, so the two solutions we got are coincide.
So if there is a solution for the initial value problem it must be included in the general solution.
For the initial value problem, $y(0)=1$ , the general solution is not defined so there is no solution for this initial value problem.   
