Where is a solution valid in an initial value problem?

[Akitirija]

The problem is from Adam's Calculus (7th Ed). It is an initial value problem, and I solved it:
$\begin{cases} y'=\frac{3+2x^{2}}{x^{2}} \\ y(-2)=1 \end{cases} \\ \implies y=-\frac{3}{x}+2x+\frac{7}{2}$

I can see that the solution is not valid for x=0, but the book says that the solutions is only valid for the interval

$(-\infty,0)$ because "that is the largest interval that contains the initial point -2 but not the point x=0.

I do not understand this. Why can x not be larger than 0?

 

[Domenico94]

Maybe because in certain points it's not differentiable. In addiction, 0 is in an open interval, that means you got to not consider that

 

[Akitirija]

Thank you for your answer, Domenico.

But is this function not differentiable for all x except 0?

And even if there are other points, should the book not mention these points instead of excluding the whole interval after 0?

 

[Domenico94]

The point in question is -2, so why should it include positive values as well? I think that's the reason

 

[Akitirija]

Thank you, Domenico! I'm not sure I entirely understand it, but maybe it becomes more apparent as I learn more :)

 

[pasmith]

The problem is from Adam's Calculus (7th Ed). It is an initial value problem, and I solved it:
$\begin{cases} y'=\frac{3+2x^{2}}{x^{2}} \\ y(-2)=1 \end{cases} \\ \implies y=-\frac{3}{x}+2x+\frac{7}{2}$

I can see that the solution is not valid for x=0, but the book says that the solutions is only valid for the interval

$(-\infty,0)$ because "that is the largest interval that contains the initial point -2 but not the point x=0.

I do not understand this. Why can x not be larger than 0?

The following function is smooth on $\mathbb{R} \setminus \{0\}$ for, and satisfies the initial value problem for, every $C \in \mathbb{R}$:
$$y(x) = \begin{cases} - \tfrac{3}{x} + 2x + \tfrac72, & x < 0,\\ - \tfrac{3}{x} + 2x + C, & x > 0. \end{cases}$$ It follows that the solution on $\mathbb{R} \setminus \{0\}$ is not unique unless you also specify the value of $y$ at some $x_0 > 0$. Usually when solving ODEs we want unique solutions, and to do that we have sometimes to restrict the domain to a neighbourhood of the point where the initial condition is to be applied.

Also, in applications, our interpretation of this solution would be that some object reaches infinity, or some quantity becomes infinite, in finite time, which suggests a problem with our model; we can't conclude that the object subsequently reappears at the opposite end of the universe, or that the quantity subsequently becomes arbitrarily large and negative.

In general, if you're antidifferentiating a function which has singularities at a finite number of points $x_1 < \dots < x_n$, then you can take different constants of integration on each of the disjoint intervals $(-\infty, x_1)$, $(x_1, x_2)$, ..., $(x_{n-1}, x_n)$, $(x_n, \infty)$, and the result will be a continuous function on $\mathbb{R} \setminus \{x_1, \dots, x_n\}$.

 

[Akitirija]

Thank you very much, pasmith! I just started studying this, so I hope I can ask you one probably very stupid question: What do you mean by "the initial condition"?