# 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?

• Assume a solution exists in a neighborhood of $t=0$. Such solution would be a solution for another initial value problem, with the same equation, but with some other initial condition $y(t_0)=y_0$, with $t_0=neq0$. Find all solutions for that new initial value problem and show that none of them pass through $(t,y)=(0,1)$. – user551819 Apr 27 '18 at 20:20
• hey could you edit the equation a little bit to clarify things, for example what is t? – shai horowitz Apr 27 '18 at 20:20
• @totoro, I think that comment is worth being posted as an answer (which is what I was doing until you out-typed me :--}) – JonathanZ Apr 27 '18 at 20:21
• BTW, if you implement totoro's method, your life will be easier if you assume a small enough interval so that $y$ is strictly positive on it, which you can do because $y(0) \gt 1$. – JonathanZ Apr 27 '18 at 20:24
• @JonathanZ Nothing prevents you from completing your solution. There are no new discoveries being made here. Only helping the next person. – user551819 Apr 27 '18 at 20:24

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 $y(0)=1$

• But wouldn't in this case you make the assumption that $t \ne 0$, because you divide by t. – Idan Aviv Apr 27 '18 at 20:31
• 'Solving it' is useful. But the subtle detail is that that process finds all solutions for which $t\neq0$. In principle there could still be other solutions passing through points with $t=0$. The problem ends by observing that such solutions will have to also be defined at some points with $t\neq0$ and by uniqueness coincide with some of the ones you found. – user551819 Apr 27 '18 at 20:33
• Ok great, could you write your comment as a solution so I could mark my question as solved. – Idan Aviv Apr 27 '18 at 20:39
• @IdanAviv Better idea. You write the solution. That way you make sure you understood the argument, plus you get the points. – user551819 Apr 27 '18 at 20:41

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.