# y'=x-(y^2)sin(y) has unique solution?

• March 25th 2009, 02:29 AM
mylestone
y'=x-(y^2)sin(y) has unique solution?
My differential equations is rusted-to-the-point-of-useless, so here's a problem, what I did, and a question about it:

I'm staring at $y'=x-(y^2)sin(y)$, with the condition that $y(0)=-2$. I need to determine whether there is a unique solution here within some sufficiently small interval about zero. So, I put the y-relevant things on the left, the x-relevant things on the right, integrate both sides and end up with $2y \sin (y) - (y^2 -2) cos (y) = \frac{x^2}{2} + C$.

Now I'm not even sure I did that properly, but if I pretend I did then I get confused here because if I use the point (0, -2) to find C then graph the implicit function using this C, my graph looks like ellipse-ish forms radiating vertically from the origin--I'm unclear how to interpret this. Is this my unique solution?

If, on the other hand, I enter the original differential equation "as-is" with the specified condition into my graphing tool, a single line (however strangely shaped) appears before me, suggesting more directly (to my inept intuition, at any rate) an affirmative "Yes, there IS a unique solution, and you're staring at it, buddy".

Methinks a wholly analytic solution would show me the link I'm missing between the two, but I'm unsure how that goes. Maybe I've just been up too long to sort it out solo, but any clarification would be greatly appreciated.

• March 28th 2009, 08:16 AM
TheEmptySet
Quote:

Originally Posted by mylestone
My differential equations is rusted-to-the-point-of-useless, so here's a problem, what I did, and a question about it:

I'm staring at $y'=x-(y^2)sin(y)$, with the condition that $y(0)=-2$. I need to determine whether there is a unique solution here within some sufficiently small interval about zero. So, I put the y-relevant things on the left, the x-relevant things on the right, integrate both sides and end up with $2y \sin (y) - (y^2 -2) cos (y) = \frac{x^2}{2} + C$.

Now I'm not even sure I did that properly, but if I pretend I did then I get confused here because if I use the point (0, -2) to find C then graph the implicit function using this C, my graph looks like ellipse-ish forms radiating vertically from the origin--I'm unclear how to interpret this. Is this my unique solution?

If, on the other hand, I enter the original differential equation "as-is" with the specified condition into my graphing tool, a single line (however strangely shaped) appears before me, suggesting more directly (to my inept intuition, at any rate) an affirmative "Yes, there IS a unique solution, and you're staring at it, buddy".

Methinks a wholly analytic solution would show me the link I'm missing between the two, but I'm unsure how that goes. Maybe I've just been up too long to sort it out solo, but any clarification would be greatly appreciated.

Unfortuatly this equation is not seperable and the above solution is not correct.

Existance and uniqueness can be determined however.

First we need to put the function in the form

$\frac{dy}{dx}=f(x,y)$ this is how the equation is given

$y'=f(x,y)=x-(y^2)\sin(y)$

Now if the above is continuous and its partial with repect to y is continuous on some rectangle containing you intial condition then it has a solution and it is unique.

The above is continous on all of $\mathbb{R}^2$

and its partial with repect to y is

$\frac{\partial }{\partial y}(x-(y^2\sin(y)))= -2y\sin(y)-y^2\cos(y)$

is also continuous on all of $\mathbb{R}^2$.

This tells us that a solution both exists and is unique.