Uniqueness - exact differential equation

[Pavoo]

Hi folks! This one got me in doubts...

Homework Statement

Solve IVP (Initial Value Problem): (2xy+sin(x))dx+(x^{2}+1)dy=0, y(0)=2

Is the solution unique? Motivate why!

Homework Equations

Relevant equations for solving the exact equation...

The Attempt at a Solution

I can solve this without any trouble. Since the answer is an implicit solution, I get:

F(x,y)=x^{2}y-cos(x)+y= C

Putting in the IVP value I get for y:

y(x)=\frac{1+cos(x)}{x^{2}+1}

Now, to the question, is this solution unique? Why? Why not? And how does it relate to implicit solutions (here, exact differential equation)?

What about if this IVP was a separable, or a linear differential equation?

I am thankful for any hints, because this one got me really thinking...

 

[LCKurtz]

Have you studied the existence and uniqueness theorem for the general first order equation $y'=f(x,y)$? Something like in this link?:

http://www.math.uiuc.edu/~tyson/existence.pdf

It is relevant to your question.

 

[Pavoo]

Thank you LCKurtz for the fast reply! I have studied the sheet, as I have with my coursebook, but didn't get any smarter. Here's where I don't get it:

Yes, the paper says and explains clearly why there is y(0)≠0 gets no solution. But how does this applies to this particular equation, meaning, which first gives an implicit solution? Or does explicit/implicit not matter here?

I get a solution to this exact differential equation, even though the paper says I shouldn't get one. Or am I misreading this entirely?

Thanks for your help!

 

[LCKurtz]

Thank you LCKurtz for the fast reply! I have studied the sheet, as I have with my coursebook, but didn't get any smarter. Here's where I don't get it:

Yes, the paper says and explains clearly why there is y(0)≠0 gets no solution. But how does this applies to this particular equation, meaning, which first gives an implicit solution? Or does explicit/implicit not matter here?

I get a solution to this exact differential equation, even though the paper says I shouldn't get one. Or am I misreading this entirely?

Thanks for your help!

Have you figured out what $f(x,y)$ is in your equation and checked the hypotheses of the uniqueness theorem against it?

 

[Pavoo]

dF/dy=\frac{-2x}{x^{2}+1}

So you are saying that the "test" against the hypotheses of uniqueness states the fact, independent whether the differential equation (IVP) has any solution - as it does in this case?

 

[LCKurtz]

dF/dy=\frac{-2x}{x^{2}+1}

So you are saying that the "test" against the hypotheses of uniqueness states the fact, independent whether the differential equation (IVP) has any solution - as it does in this case?

That link I gave you has both an existence and uniqueness theorem. If a function satisfies the hypotheses of the uniqueness theorem, doesn't it automatically satisfy those for the existence theorem? So if your $f(x,y)$ satisfies the hypotheses of the uniqueness theorem, then it has a solution and it is unique. Does your $f(x,y)$ satisfy the hypotheses? That is your only question. You hardly need the existence theorem given that you already have a solution.

 

[Pavoo]

Thanks for your effort!

However I am still confused with this, especially with implicit solutions.

As a last try, let me rephrase my question:

The paper says: "no solution if x₀=0 and y₀≠0." . But here, I do have a solution for the differential equation. Maybe that last summary is irrelevant if I have this single solution.

 

[LCKurtz]

Thanks for your effort!

However I am still confused with this, especially with implicit solutions.

As a last try, let me rephrase my question:

The paper says: "no solution if x₀=0 and y₀≠0." . But here, I do have a solution for the differential equation. Maybe that last summary is irrelevant if I have this single solution.

They are talking about their equation. You need to check the theorems for your equation. And whether or not a solution exists has nothing to do with whether you have an explicit or implicit formula.

 

[Pavoo]

Whoops!
So let me try it out in this case, and please correct me if I am wrong:

dy/dx=-\frac{(2xy+sin(x))}{x^{2}+1}

is 0 for x=0, y=2.

dF/dy=-\frac{(2x)}{x^{2}+1}

is 0 for x=0, y=2. However, this still means that the solution is unique, because both functions are continuous and defined near x=0, y=2. Meaning, the solution exists, and it is unique.

 

[HallsofIvy]

Yes, the first thing you should have done, after writing the equation as
\frac{dy}{dx}= \frac{2xy+ sin(x)}{x^2+ 1}
is recognize that the denominator is never 0 and so the function is continuous for all x. Since the derivative, using the quotient rule will just have the square of x^2+ 1 in the denominator, it is still never 0 and so the derivative is continuous for all x.

 

[Pavoo]

Thanks to both of you for your hints and guidance!