When is a Constant a Solution to a Differential Equation

Click For Summary

Discussion Overview

The discussion revolves around the conditions under which a constant, specifically the solution \(x=0\), can be considered a solution to a differential equation. Participants explore the implications of this in the context of a specific differential equation and the nature of singular solutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why \(x=0\) is a solution to the differential equation \((x^2-3y^2)dx+(2xy)dy=0\), suggesting that plugging \(x=0\) into the equation yields a form that results in no change.
  • Another participant asserts that if \(x=0\), both \(x\) and \(dx\) are zero, leading to both terms in the equation being zero.
  • A follow-up question arises regarding the nature of \(dx\) being zero, with a participant stating that \(x\) is constant in this case.
  • There is a discussion about whether this reasoning holds for other constant values of \(x\), such as \(x=1\), with participants noting that the \(dy\) term would not be zero in such cases.
  • One participant mentions that when solving the differential equation, \(x=0\) is typically omitted from the general solution due to the division by \(x\), which raises questions about the restrictions on solutions.
  • Another participant clarifies that \(x=0\) was not omitted from the general solution according to their textbook, prompting further exploration of the nature of singular solutions.
  • A later reply references the need to check for singular solutions when dividing by terms that could equal zero, emphasizing the importance of writing restrictions in the context of differential equations.
  • Participants express uncertainty about the implications of their reasoning and invite corrections, indicating a collaborative learning environment.

Areas of Agreement / Disagreement

Participants express differing views on whether \(x=0\) should be considered a solution and the implications of omitting it from the general solution. There is no consensus on the nature of singular solutions or the restrictions that apply.

Contextual Notes

Participants highlight the importance of understanding the conditions under which certain solutions are valid, particularly in relation to operations like division that may exclude specific values from the solution set.

Drakkith
Mentor
Messages
23,205
Reaction score
7,687
I've run across several instances while doing homework where a question will have two solutions. One will be an equation, and the 2nd will be a constant (usually zero). I can't figure out why this constant is a solution though.

For example, take the following differential equation: $$(x^2-3y^2)dx+(2xy)dy=0$$

The solutions to this differential equation turn out to be $$y=±\sqrt{x^2+Cx^3}$$
and$$x=0$$

Why is ##x=0## a solution? The only thing I can think of, and this is mostly a guess, is that plugging ##x=0## into the original equations yields $$(-3y^2)dx=0$$
Is it that ##y## is taken to be a constant here and since there's no ##x## variable there is no change?
 
Physics news on Phys.org
If x=0 it does not matter what y is because both x and dx are zero and so both terms contain factors that are zero.
 
Orodruin said:
because both x and dx are zero

Why is ##dx## zero?
 
Because x is constant.
 
I'm assuming that's not true for, say, ##x=1##. Is that because the ##dy## term isn't zero then?
 
Drakkith said:
I'm assuming that's not true for, say, ##x=1##. Is that because the ##dy## term isn't zero then?
##dx## would still be zero if ##x## was constantly equal to one. Indeed, the ##dy## term is not zero then.
 
Alright. Thanks Oro.
 
When you solved the differential equation you had to divide both sides by x, since x ≠ 0, x=0 will be omitted from your general solution by default. Remember to write your restrictions, because it's impossible to get singular solutions by varying the parameters of your general solution as far as I know.
 
VuIcan said:
When you solved the differential equation you had to divide both sides by x, since x ≠ 0, x=0 will be omitted from your general solution by default. Remember to write your restrictions, because it's impossible to get singular solutions by varying the parameters of your general solution as far as I know.

What restrictions? X=0 was not omitted from the general solution as far as the book is concerned. That was the point of my question.
 
  • #10
Drakkith said:
What restrictions? X=0 was not omitted from the general solution as far as the book is concerned. That was the point of my question.
I misunderstood your question, my apologies.
 
  • #11
VuIcan said:
I misunderstood your question, my apologies.

Well, looking back at the question, I now have another question. The equation can easily be rearranged into ##\frac{dy}{dx}=\frac{3y^2-x^2}{2xy}##
In this form you certainly have a problem when ##x=0##, right?
 
  • #12
Drakkith said:
Well, looking back at the question, I now have another question. The equation can easily be rearranged into ##\frac{dy}{dx}=\frac{3y^2-x^2}{2xy}##
In this form you certainly have a problem when ##x=0##, right?

I'm self-taught on this topic, so take what I say with a grain of salt. But here's what I was referring to:

https://en.wikipedia.org/wiki/Singular_solution

"Therefore, when one is solving a differential equation and using division one must check what happens if the term is equal to zero, and whether it leads to a singular solution."

This is why I urged you to write your restrictions. I assumed you were wondering why your textbook had constant values as solutions. It's because those constant values are omitted on occasion when you perform certain operations(division in this case) from the domains of all the particular solutions obtainable via varying the constant in your general solution.

Again, I'm self taught (still learning in fact) so take what I say with a grain of salt and please do correct me if I'm mistaken. Thanks in advance.
 

Similar threads

Undergrad Why Lagrange’s method of solving Pp + Qq=R works?
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Solving Differential Equation Using Reduction of Order
  • · Replies 2 ·
Replies
2
Views
4K
Undergrad How to characterise solutions to an unsolved equation
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad General solution vs particular solution
  • · Replies 52 ·
2
Replies
52
Views
9K
Undergrad Expressing a differential equation into a different format
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Integral-form change of variable in differential equation
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Correct Upper/Lower limits for continuation of solutions for ODE
  • · Replies 0 ·
Replies
0
Views
3K
MHB Solve Differential Eq: xe^-1/(k+e^-1) for x, k, t
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Direction Fields and Isoclines
  • · Replies 1 ·
Replies
1
Views
1K
Graduate Differential equation and Appell polynomials
  • · Replies 1 ·
Replies
1
Views
4K