# Why are the solutions of differential equations given with restricted domains?

I solved $y^2=y'$ with the initial condition that $y(0)=1$, and I got $\displaystyle y(x)=\frac{1}{1-x}$.

My book says it's only the portion of this function from $-\infty$ to $1$ that counts.

I don't understand why?

Now, if the initial condition were $y(2)=-1$, would that make it the portion from $1$ to $+\infty$?

What if both initial conditions were given?

• There's a singularity at $x = 1$, so your system blows up and you can't get past it from $x = 0$. – Qiaochu Yuan Jun 1 '11 at 22:11
• I want to add that I know we are avoiding the singularity at one, but I don't know WHY it is important to do this. – Noteventhetutorknows Jun 1 '11 at 22:16
• You can only start from $x = 0$ and work outward. There's no physical way to get past the singularity, so there's no reason to suppose that $y$ is defined past it. Mathematically, you only know that if $y$ is defined to the left of $1$ then it has the form $y = \frac{1}{C-x}$ (or $y = 0$) there, but you have no way of knowing what $C$ is. – Qiaochu Yuan Jun 1 '11 at 22:18
• Oh that makes sense, so I'd need an initial condition for every interval where it is possible to choose from multiple branches in the family of solution curves. I could take (0,1) and (4, -1) as initial conditions and get a price wise function from 1/(1-x). And 1/(3-x)... For example. Neat. – Noteventhetutorknows Jun 1 '11 at 22:24
• If this is a physical problem where $x$ is time, then there's a blowup after finite time and it doesn't make sense to try to continue past the singularity. But in some contexts the "time" variable is complex, and then you can go around the pole by analytic continuation, so that $y(x)=1/(1-x)$ would be the unique solution in the whole complex plane (minus the point $x=1$). – Hans Lundmark Jun 2 '11 at 7:21

Let us look at a more familiar problem. We want to find a function $y$ such that $$\frac{dy}{dt}=-\frac{1}{(t-3)^2} \qquad \text{and} \qquad y(4)=17$$

Integrate. If we do it in the familiar first year calculus way, we get $1/(t-3)+C$. To evaluate $C$, put $t=4$. Quickly we get $y(t)=-1/(t-3)+18$.

For what $t$ is this correct? I claim that it is only for $t >3$. To see why, assume that $y$ represents the displacement of a particle at time $t$. Then $\frac{dy}{dt}$ represents its velocity.

The velocity (and displacement) are not defined at time $t=3$. From the information that the velocity at any time $t\ne 3$ is $-1/(t-3)^2$, and the information that the displacement at $t=4$ is $17$, we can find the displacement at any time $t>3$.

But we cannot go backwards past the singularity at time $t=3$ to reach any conclusion about displacement at any time $t<3$.

Similarly, if we had been told the displacement at say $t=1$, we could find the displacement at any time $t<3$, but we could get no information about $t>3$.

If we want a "general" formula for the $\int\frac{-dt}{(t-3)^2}$ that holds for as wide a domain as possible, we would have to say that $y=-1/(t-3)+C_1$ for $t>3$ and $y=-1/(t-3)+C_2$ for $t<3$, where $C_1$ and $C_2$ are constants, possibly different constants.

But more usually we restrict the domain, and like in our example with the initial condition $y(4)=17$, we simply say that $y(t)=-1/(t-3)+18$ for $t>3$.

Roughly speaking the same sort of thing happens in your differential equation example. You presumably solved the equation by rewriting as $$\frac{dy}{y^2}=dx$$ then integrating, and applying the initial condition.

Look at what you got for $y$. It (and its derivative) blows up at $x=1$. So like in the integration problem I started with, initial information about $x=0$ can't give any knowledge about the function past its singularity at $x=1$.

Suppose now we take your suggested $y(2)=-1$. When solving the DE, we get $-1/y=x+C$, so $C=-1$, I hope. That gives $y=-1/(x-1)$, valid for $x>1$.

But change things to $y(2)=1$. We get $C=-3$. So the position of the singularity can be affected by the initial condition. (This cannot happen in simple integration examples.)