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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

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    $dy/dt=1/t$, $dy=dt/t$, $\int\,dy=\int\,dt/t$, where do you get that square root?2012-05-17
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    @ Gerry Myerson,see please edited2012-05-17
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    This equation is meaningless. Substitute $t=0$2012-05-17
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    then $y(0)=0$ so no problem with solution right?2012-05-17
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    @dato: With the DE $\frac{dy}{dt}=\frac{1}{y}$, you would end up with square roots of the type you mention. Can you check carefully which DE is under discussion? There is a big difference between $\frac{dy}{dt}=\frac{1}{t}$ and $\frac{dy}{dt}=\frac{1}{y}$.2012-05-17
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    last one @André Nicolas $dy/dt=1/y$ i have made typo when was posting question2012-05-17
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    I'm a bit inclined to construe the problem as follows: find a continuous function whose value at $0$ is $0$ and that satisfies the differential equation at points where the latter makes sense. And I'm not altogether sure that we must reject the idea that the value of the derivative and also that of $1/y$ is $\infty$ or $-\infty$ at $t=0$. We can certainly say, not only that certain limits as $t\to0$ have certain values, but also that there is indeed a tangent line to the curve at $t=0$, and it is vertical.2012-05-17
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    While it is tempting to brush off the misbehavior at zero as not being enough to worry about, it would be a mistake: in real problems, such details often turn out to be important, and it's usually worth fully considering them.2012-05-18

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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.

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    thanks too much,i have understood everything now2012-05-17
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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