Undergrad When is a Constant a Solution to a Differential Equation

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SUMMARY

The discussion centers on the nature of constant solutions in differential equations, specifically examining the equation $$(x^2-3y^2)dx+(2xy)dy=0$$. The solutions include both a general form $$y=±\sqrt{x^2+Cx^3}$$ and the constant solution $$x=0$$. Participants clarify that while $$x=0$$ is a valid solution, it is often omitted in general solutions due to division by zero when rearranging the equation. The importance of recognizing singular solutions and writing restrictions in differential equations is emphasized.

PREREQUISITES
  • Understanding of differential equations and their solutions
  • Familiarity with singular solutions in mathematics
  • Knowledge of the implications of division by zero in equations
  • Basic calculus concepts, including derivatives and integrals
NEXT STEPS
  • Research the concept of singular solutions in differential equations
  • Study the implications of division by zero in mathematical equations
  • Learn about the methods for solving differential equations, including separation of variables
  • Explore the role of constant solutions in various types of differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as self-taught learners seeking to deepen their understanding of solution types and restrictions in mathematical contexts.

Drakkith
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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?
 
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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.
 

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