# Thread: the interval for which the solution to DE makes sense

1. ## the interval for which the solution to DE makes sense

Here's an exercise from Tenebaum & Pollard's book:

Differential equation: $x \frac{dy}{dx} = 2y$
Solution: $y = x^{2}$
Give the interval on which the solution makes sense.
I know how to show that the bottom one is the solution of the top equation, what I don't understand is why the interval on which the solution works is not:
$-\infty < x < \infty$

$x \neq 0$ as given the by author.

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Trying to understand this intuitively, I'm thinking that anything multiplied by zero is zero. So when $x = 0$, anything substituted on the left hand side gives zero. However, if $y = x^{2}$ is the solution, when $x = 0$, the RHS is always zero.

I don't see where the restriction $x \neq 0$ comes from.

Thanks for your help!

2. "The" solution $y = x^2$ is not entirely correct. Every ODE has an infinite family of solutions, usually parameterized by a constant of integration:

$y = Cx^2$

is a more general solution. Even $C=0$ works; i.e., the function $y \equiv 0$ satisfies the differential equation from the book.

To try to understand the domain where the solution makes sense intuitively, try thinking of solutions as curves (i.e., integral curves) in the $xy$-plane, and think about the uniqueness of solutions. Given any point $(x_0, y_0)$ where $x_0<0$, you can can find a constant $C_0$ such that the curve

$y(x) = C_0x^2$

passes through the point $(x_0, y_0)$. However what happens at $x=0$? For any (other) constant $C_1$, the curve $z(x)=C_1x^2$, $x > 0$, is also a solution to the DE, and it meets the solution $y(x)$ at the origin. Uniqueness is violated.

For any point in the domain $(-\infty, 0) \times \mathbf{R}$, there is a unique solution of the DE passing through that point that is defined for all $x \in (-\infty,0)$. The same is true for the domain $(0, \infty)\times \mathbf{R}$. This is why the authors exclude $x=0$.

Hope that helps!

Jerry