Missing Solutions Homework: Find 1-Param Family for DE

• roam
In summary: I'm not sure if it is a solution or notno i mean that if it was a solution, we might...I'm not sure if it is a solution or not
roam

Homework Statement

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.

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?

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

sunjin09 said:
There could be a negative sign in front of the square root, with an independent constant c(or k).

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.

hi roam!
roam said:
How does that show that there are any missing solutions?

sunjin09 is correct

if y is a solution, then obviously -y is also
Furthermore, there is no negative sign in front of the squre root in the general solution that I've found.

that's because you deliberately threw it away when you did the following step
roam said:
$\frac{y^2}{2} = \frac{1}{4} (2t \ sin (2t)+ cos(2t)) + c$

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

roam
tiny-tim said:
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

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)?

roam said:
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)?

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

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

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)

roam
tiny-tim said:
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)

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

Ray Vickson said:
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.

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

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

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

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).

roam said:
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?

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

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

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 ±"
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).

looks ok

roam
tiny-tim said:
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

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

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Ray Vickson said:
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

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?

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

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

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

Who says there are missing solutions?

RGV

Ray Vickson said:
Who says there are missing solutions?

RGV

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?

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?

roam said:
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?

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

roam

1. What is a "1-Param Family" in the context of solving a differential equation?

A 1-Param Family in the context of solving a differential equation refers to a set of equations that can be derived from a single general solution by varying a single parameter. This allows for a range of solutions to the same differential equation, each with a unique value for the parameter.

2. How do you find a 1-Param Family for a given differential equation?

To find a 1-Param Family for a given differential equation, you must first solve the differential equation by finding a general solution. Then, introduce a new parameter into the general solution and vary its value to generate a range of solutions.

3. What is the purpose of finding a 1-Param Family for a differential equation?

The purpose of finding a 1-Param Family for a differential equation is to generate a range of solutions that satisfy the same equation but have different characteristics. This can be useful in modeling real-world situations where there may be multiple possible solutions to a problem.

4. Can a 1-Param Family be used to solve any type of differential equation?

No, a 1-Param Family can only be used to solve certain types of differential equations, specifically those that are separable or can be solved by integrating factors. It cannot be used for more complex types of differential equations, such as partial differential equations.

5. How does finding a 1-Param Family differ from finding a particular solution to a differential equation?

Finding a 1-Param Family involves finding a range of solutions to a differential equation by introducing a new parameter, while finding a particular solution involves finding a single solution that satisfies specific initial conditions for the differential equation. A 1-Param Family is more general and allows for a variety of solutions, while a particular solution is more specific and unique.

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