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I'm working through Exercise $3$ in Tenenbaum and Pollard's ODE book. The question is about determining whether functions are solutions to differential equations, and to state the interval where the differential equation makes sense. The part I'm stuck on is:

$$\text{(e) }xy'=2y, \text{ } y=x^2$$

The actual calculation to demonstrate that this is true is simple enough.

$$y'=2x\rightarrow xy'=2y\rightarrow 2x^2=2(x^2)\text{ }$$

However, the solution states that the interval where it makes sense is $x\neq 0$, which is confusing to me. I don't understand why the interval isn't $-\infty<x<\infty$, as neither the function nor the differential equation have any discontinuties at $x=0$. Am I missing something, or is this a mistake in the solutions? Any input would be greatly appreciated.

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  • $\begingroup$ The differential equation has a singularity at $x = 0$. Therefore, it is defined for any interval which does not contain $0$. $\endgroup$ Jul 6, 2019 at 4:48
  • $\begingroup$ Ok, so I suppose I don't understand why the differential equation has a singularity at $x=0$. I'm sorry if it's really basic, but would you mind explaining the intuition behind this? $\endgroup$
    – scoopfaze
    Jul 6, 2019 at 4:51
  • $\begingroup$ @scoopfaze I believe the singularity comes from when you isolate $y'$ by itself to get $y' = \frac{2y}{x}$. At $x = 0$, the right side is undefined. $\endgroup$ Jul 6, 2019 at 5:08
  • $\begingroup$ @JohnOmielan Yea of course, I should've seen that. Thank you! $\endgroup$
    – scoopfaze
    Jul 6, 2019 at 5:10

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In any differential equation, the derivative must be well-defined for all values for which the equation is expected to hold. In your case, you have

$$xy' = 2y \tag{1}\label{eq1}$$

The singularity you mention comes from when you isolate $y'$ by itself to get

$$y' = \frac{2y}{x} \tag{2}\label{eq2}$$

As Aniruddha Deshmukh stated in a question comment, the right side is undefined at $x = 0$, so the derivative is not properly defined there and, thus, the solution is not either.

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  • $\begingroup$ Thanks! For ODEs with degree $>1$, I assume you would need to check all derivatives, or would it be the one with the largest degree? $\endgroup$
    – scoopfaze
    Jul 6, 2019 at 5:21
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    $\begingroup$ @scoopfaze You're welcome. With ODEs with degree $\gt 1$, consider a derivative at any degree, say $n$, which is not defined. For the next higher derivative, i.e., at $n + 1$, note the limit definition for derivatives gives that $f^{n+1}(x) = \lim_{h \to 0} \frac{f^{\, n}(x + h) - f^{\, n}(x)}{h}$, where the power is the degree of the derivative. As you can see, you need $f^{\, n}(x)$ to be well defined for $f^{n+1}(x)$ to exist. Thus, the higher degree derivatives can't exist if any lower one doesn't. Note it's possible for an equation to have the largest degree be well-defined, but some ... $\endgroup$ Jul 6, 2019 at 5:29
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    $\begingroup$ @scoopfaze (cont.) lower-level ones to not be, so you should in theory always check all of the degrees of the derivatives. I don't offhand see how this situation can arise in just one equation (although I don't know to prove it can't), but it's certainly quite possible if you're dealing with multiple equations. $\endgroup$ Jul 6, 2019 at 5:32
  • $\begingroup$ That makes a lot of sense. Thanks again. $\endgroup$
    – scoopfaze
    Jul 6, 2019 at 5:34

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