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I feel some thing unclear in finding the singular solution ($y=0$, by $p$-discriminant method) for the following differential equation

$$f \left(x,y,y'\right)=xy'+2y=0,$$

while the general solution found as $$y=\frac{C}{x^2},$$ where $C$ is an arbitrary constant. See that $f_{y'}=0$ is not sensible for the general solution since $x \neq 0$. On what conditions we can ensure the existence of singular solutions? Is it completely independent of the general solution. How perfectly we can define the singular solution of an ordinary differential equation and the necessity of existence?

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$y$ need only be differentiable where $x\neq0,$ so the domain is taken to exclude $0.$ In general, if $a(x)y'(x)=f(x,y(x)),$ then you can expect the solution to be singular whenever $a(x)=0.$ The singularities sometimes are removable, under very restricting circumstances, but this is not to be expected in general. In this case, the domain considered is $\mathbb{R}\setminus\{0\},$ not $\mathbb{R}.$ The solution $y(x)=C/x^2$ works everywhere in the domain for every real $C.$ However, the one for $C=0$ is the only one that can be continuously extended to $\mathbb{R},$ yielding $y=0,$ and this last solution is the only solution differentiable in $\mathbb{R}.$

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  • $\begingroup$ Sir, I am not getting how you connected the singular solutions mentioned in the answer and the solutions of ODE on which each point is an initial data making failure in the uniqueness. $\endgroup$
    – Riaz
    Commented Mar 18, 2022 at 16:53
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    $\begingroup$ @RIYASUDHEENTK I am not sure I understand what you are asking. I never said anything about uniqueness, except in my last sentence, and your post says nothing about uniqueness either. $\endgroup$
    – Angel
    Commented Mar 18, 2022 at 17:13
  • $\begingroup$ Actually I couldn't find a perfect definition for singular solutions of an ODE from the literature. Some definitions says they are solutions $y=y(t)$ of the ODE on which each point $(t_0,y_0)$ as an initial data makes failure in the uniqueness. $\endgroup$
    – Riaz
    Commented Mar 19, 2022 at 0:56
  • $\begingroup$ Some of the definitions asserts that singular solutions cannot be obtained from the general solution as well. $\endgroup$
    – Riaz
    Commented Mar 19, 2022 at 1:16

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