# DE Missing Solutions

1. Mar 14, 2012

### roam

1. The problem statement, all variables and given/known data

Find a one-parameter family of solutions to the differential equation:

$\frac{dy}{dt} = \frac{t \ cos 2t}{y}$

Are there any solutions to the differential equation that are missing from the set of solutions you found? Explain.

3. The attempt at a solution

I used the separation of variables to find the solution as follows:

$\int ydy = \int t \ cos (2t) dt$

$\frac{y^2}{2} = \frac{1}{4} (2t \ sin (2t)+ cos(2t)) + c$

$\therefore y = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}$

And this is defined as long as there is no negative under the square root. So, how do I know whether there are any missing solutions or not? And how do I identify them?

2. Mar 14, 2012

### sunjin09

There could be a negative sign in front of the square root, with an independent constant c(or k).

3. Mar 15, 2012

### roam

How does that show that there are any missing solutions? Furthermore, there is no negative sign in front of the squre root in the general solution that I've found.

4. Mar 15, 2012

### tiny-tim

hi roam!
sunjin09 is correct

if y is a solution, then obviously -y is also
that's because you deliberately threw it away when you did the following step

5. Mar 15, 2012

### roam

Thank you! So the correct general solution to the DE must have been:

$\therefore y = \pm \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}$ .......(1)

But this doesn't answer the second part of the question: Are there any solutions to the differential equation that are missing from the set of solutions you found?

Are there any solutions to the DE that are missing from the set (1)?

6. Mar 15, 2012

### Ray Vickson

Your "solution" is not a solution at all; a solution will not have a $\pm$ in it. You need to pick either $$y_1(t) = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_1} \text{ or } y_2(t) = -\sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k_2}.$$ (If you don't believe me, try plugging in you written solution into a program such as Maple and see whether or not the soflware can handle it.) That being said, the question becomes: are there any solutions different from $y_1 \text{ or } y_2$? You need to use some theorems, for example, related to the IVP, where in addition to the DE you also specify y(0) for example.

RGV

7. Mar 15, 2012

### tiny-tim

well, i don't see how we could have thrown any away …

(the only dangerous thing we did was to multiply both sides by y, and "y = 0 for all t" obviously isn't a solution in this case)

8. Mar 15, 2012

### roam

So, do you mean that y(t)=0 is a missing solution?

So we must when we are solving a DE and have a situation like y2=..., we must always take the positive value?

9. Mar 15, 2012

### tiny-tim

no i mean that if it was a solution, we might have missed it

so we check, we find it isn't, and we relax again

10. Mar 16, 2012

### roam

Hi tiny tim,

I have two questions:

1. If using the "±" symbol is wrong, then what is the correct notation for representing the solution to this DE?

2. So, there are absolutely no missing solutions, and my answer is the general solution and we can solve all initial-value problems with solutions of this form? (Personally I can't think of any counter examples, but I'm not sure).

11. Mar 17, 2012

### Ray Vickson

Why would you ask me that, when I very clearly wrote down TWO solutions?

Anyway, if you look at the IVP and have y(0) > 0, then you must choose the solution y1 (with the + sign); if y(0) < 0 you must choose y2 (with the - sign). Then the issue is: are there any other solutions to your IVP? You can use theorems about that. Google "uniqueness theorems for differential equations".

RGV

Last edited: Mar 17, 2012
12. Mar 17, 2012

### tiny-tim

hi roam!
i didn't say it was wrong!

if your professor doesn't like it, don't do it

if your professor doesn't mind it, do do it

it's a convenient shorthand …

when we write "the general solution is x = Acoskt + C", we don't feel the need to add "for any constant C"

when we write "the general solution is x = -b ±√(b2 - 4c), we don't feel the need to add "for either value of ±"
looks ok

13. Mar 17, 2012

### Ray Vickson

The point I was trying to get across to him was that y = ± f(t) is _not_ a solution; it is TWO DIFFERENT solutions, but written in shorthand. As I suggested to him, try submitting his "solution" y = ± f(t) to Maple, Mathematica or Matlab to test it. The computer would choke. I know, I know that we sometimes say "the root is a ± b" and similar language, but *if one does not fully understand this one should avoid it*! It is really a shorthand form of the statement that there are two roots, a + b and a - b. It is perfectly acceptable to say "the roots are a ± b"; this emphasizes roots (plural) and is absolutely correct in all respects.

RGV

Last edited: Mar 17, 2012
14. Mar 17, 2012

### roam

By the Uniqueness Theorem, to have a unique solution both f(t, y) and ∂f/∂y must be continious at a point.

So $\frac{\partial f}{\partial y} = - \frac{t \ cos \ 2t}{y^2}$

When y=0, the partial derivative fails to exist, this means that the uniqueness theorem doesn't tell us anything about solutions of an IVP that have the form y(t0)=0. For example at the point y(0)=0, neither f(t, y) or ∂f/∂y exist. But how can we use this to prove that y(0)=0 is another solution to the IVP?

I'm confused because the satisfaction of the theorem simply guarantees a uniques solution, but failure of the theorem doesn't prove multiple solutions (inconclusive). So how can we use this theorem?

15. Mar 17, 2012

### Ray Vickson

So, if y(0) is not zero you have a unique solution; in fact, you can see from the DE itself that y(t) is strictly increasing if y(0) > 0 and is strictly decreasing if y(0) < 0, so y(t) never reaches zero. The case y(0) = 0 is pathological, and I don't think you can say the DE even means anything right at the point t = 0.

RGV

16. Mar 18, 2012

### roam

Right, so how else can I find the missing solutions?

17. Mar 18, 2012

### Ray Vickson

Who says there are missing solutions?

RGV

18. Mar 18, 2012

### roam

So f(t, y) is not defined at y=0, but it is continious everywhere else and ∂f/∂y is continious everywhere, and by the Uniqueness Theorem the solution to every initial-value problem is unique. Is this a sufficient explanation that there are no missing solutions?

19. Mar 20, 2012

### roam

The Uniqueness Theorem says if f(t,y) and ∂f/∂y are BOTH continuous functions of t and y for all t and y, then there is a unique solution to the DE for any particular choice of the initial condition.

To show that there are no missing solutions I must show thate there are NO values of t and y at which either f(t,y) and ∂f/∂y is discontinuous. BUT at the point y=0 the function f(t,y) is not defined/has a discontinuity. So one of the hypotheses of the theorem is violated. Doesn't this mean there is a missing solution?

20. Mar 20, 2012

### Ray Vickson

No. The conditions you cite are *sufficient* for uniqueness, not necessary. In other words, even if the conditions are violated that does not mean that uniqueness fails; in some examples it will, and in other examples it won't.

You KNOW that the DE is meaningless at y = 0, so you could say that it must hold if y ≠ 0. That does not mean that y(t)=0 is impossible; it just means that the DE fails at such a point. For example, the solution $y(t) = \sqrt{t \sin(2t) + \cos(t)/2 - 1/2}$
gives $y(t) \rightarrow 0 \text{ as } t \rightarrow 0,$ and $y(0) = 0,$ so y(t) is continuous at t = 0. The DE holds for all t > 0 but not at t = 0 itself.

RGV