From Tenenbaum's Ordinary Differential Equations", ch. 1, Exercise 2e:

Prove that the function is a solution of the differential equation, and state the common interval for which solution and differential equation make sense.

Differential Equation: $xy^{\prime} = 2y$

Function: $y=x^{2}$

Proving that $y=x^{2}$ satisfies the differential is fine, I got that part. The book says the set on which $y=x^{2}$ is a solution is: $x \neq 0$.

I see that if you rearrange the differential to $y^{\prime}=\frac{2y}{x}$, why x cannot equal zero. But how does one know to rearrange the differential relation like this in the first place?

IOW, why is the division by x introduced into the original relation in order to get $x\neq 0$?

And what if this were a higher order differential, of order n for example: $F(x,y,y^{\prime},\cdots,y^{(n)})=0$. Is the rule/convention that one should express $y^{(n)}$ (the highest order differential) as an explicit function, with $x,y,y^{\prime},\cdots,y^{(n-1)}$ as independent variables?

  • $\begingroup$ This is a common method of solving first order differential equations which are referred to as "separable." We can separate the variables (move them to different sides) and integrate both sides to determine what $y$ was originally. $\endgroup$ Commented Dec 30, 2015 at 23:47
  • $\begingroup$ I get that, but this book that I'm working through hasn't covered separable variables yet (I have covered it in the introduction to DE's in Stewart's Calculus, however). So I was hoping I wouldn't have to use that method to be able to determine the interval on which this function is a solution for the DE. $\endgroup$
    – mathiness
    Commented Dec 31, 2015 at 0:01
  • $\begingroup$ $x=0$ seems fine to me, as the derivative and value of the function at this point are all $0$ so we get $0=0$. I think your book is telling you to be careful when you divide by a variable because it often causes issues if that variable is zero. Not in this case. $\endgroup$ Commented Dec 31, 2015 at 0:23
  • 3
    $\begingroup$ I would be inclined to say that the book is wrong on this point. We certainly have that $y=x^2$ satisfy $xy'=2y$. Algebraically, there is no doubt on this point. At $x=0$ one might note that the equation behaves badly in some sense (e.g. doesn't yield a unique solution), but this is immaterial to the fact that $y=x^2$ is a solution. $\endgroup$ Commented Jan 2, 2016 at 0:56
  • $\begingroup$ It is probably just a convention. The function $y(x)=x^2$ solves the given differential relation for all $x\in\mathbb{R}$. However, that equation is singular for $x=0$. So maybe the author wants to treat singular points in a distinguished manner for some purpose. $\endgroup$ Commented Jan 2, 2016 at 0:59

1 Answer 1


The general solution is actually

$$\int \frac{1}{y}dy=2\int\frac{1}{x}dx\\ \ln y = 2\ln x +C\\ y=e^Ce^{2\ln x}=Cx^2$$

Assuming $x_0\neq0$, you can choose any initial point $(x_0,y_0)$ and find a unique solution of this form by letting $$C=\frac{y_0}{x_0^2}$$ For example, the solution of the differential equation that passes through $(4,2)$ will be $y=\frac{x^2}{8}$.

But if $x_0=0$, then we have $$xy^\prime=2y\\ 0=2y\\ y=0$$ So any initial value $(0,y_0)$ where $y_0\neq 0$ cannot be a solution of the differential equation. Also, notice that every solution of the form $y=Cx^2$ must pass through $(0,0)$, so you cannot find a unique solution at this point. In other words, if the differential equation describes the path of a particle, placing the particle initially at $(0,0)$ could lead to an infinite amount of trajectories. So the solution makes no sense at $x_0=0$.

There is no rule or convention about expressing higher order differential equations in the way you suggest. However, the Wronskian can be used to check if the general solution of a differential equation is valid for certain initial values. Your book should cover this in later chapters.

  • $\begingroup$ Pedantic point: the separation of variables procedure should not technically detect the solution $y \equiv 0$. Also, writing the antiderivative as $\ln(y)$ rather than $\ln(|y|)$ means that you should not technically detect the solutions $Cx^2$ for $C<0$, either. $\endgroup$
    – Ian
    Commented Jan 2, 2016 at 0:55
  • $\begingroup$ "The general solution is actually..." not at all what you write. It happens that $xy'=2y$, $y(0)=0$, is a classical example of a differential equation with nonunique solutions, for example each $y$ defined by $y(x)=ax^2$ for $x>0$, $y(0)=0$, and $y(x)=bx^2$ for $x<0$, solves it. $\endgroup$
    – Did
    Commented Jan 2, 2016 at 1:16
  • $\begingroup$ The problem from the book doesn't ask you to find the general solution, and the book at this point hasn't covered how to solve differential equations. This chapter is just covering what constitutes a solution, and they provided a differential equation and a function that satisfies the equation, and they asked you to find which interval it's a solution on. So while your method would possibly work to explain, I don't think the book expects the reader to be able to use this method. $\endgroup$
    – mathiness
    Commented Jan 2, 2016 at 3:17

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