# I When is a Constant a Solution to a Differential Equation

1. Jun 20, 2017

### Staff: Mentor

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?

2. Jun 21, 2017

### Orodruin

Staff Emeritus
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.

3. Jun 21, 2017

### Staff: Mentor

Why is $dx$ zero?

4. Jun 21, 2017

### Orodruin

Staff Emeritus
Because x is constant.

5. Jun 21, 2017

### Staff: Mentor

I'm assuming that's not true for, say, $x=1$. Is that because the $dy$ term isn't zero then?

6. Jun 21, 2017

### Orodruin

Staff Emeritus
$dx$ would still be zero if $x$ was constantly equal to one. Indeed, the $dy$ term is not zero then.

7. Jun 21, 2017

### Staff: Mentor

Alright. Thanks Oro.

8. Jun 27, 2017

### VuIcan

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.

9. Jun 27, 2017

### Staff: Mentor

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. Jun 27, 2017

### VuIcan

I misunderstood your question, my apologies.

11. Jun 27, 2017

### Staff: Mentor

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. Jun 27, 2017

### VuIcan

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.