# Thread: Crazy Domain Problem from an AP Exam

1. ## Crazy Domain Problem from an AP Exam

I'll post the entire problem. My math class and I couldn't figure it out. My teacher had a helluva hard time trying to explain it to us. I just want to see if anyone can understand this problem and explain it better to me. I still don't understand it
Here it goes:

"Consider the differential equation dy/dx=(1+y)/x, where x does not equal 0.

(a) On the axis provided, sketch a slope field for the given differential equation at the eight points indicated. (you can just skip this one and get to the hard question)

(b) Find the particular solution y=f(x) to the differential equation with the initial condition f(-1)=1 and state its domain."

So the last part of part b is the problem I am really asking for you to solve, although the first part of (b) as well as (a) will help.

2. Originally Posted by Skerven
I'll post the entire problem. My math class and I couldn't figure it out. My teacher had a helluva hard time trying to explain it to us. I just want to see if anyone can understand this problem and explain it better to me. I still don't understand it
Here it goes:

"Consider the differential equation dy/dx=(1+y)/x, where x does not equal 0.

(a) On the axis provided, sketch a slope field for the given differential equation at the eight points indicated. (you can just skip this one and get to the hard question)

(b) Find the particular solution y=f(x) to the differential equation with the initial condition f(-1)=1 and state its domain."

So the last part of part b is the problem I am really asking for you to solve, although the first part of (b) as well as (a) will help.
dy/dx = (1 + y)/x

We need to seperate the variables: divide out the (1 + y) and move the dx to the RHS:

dy/(1 + y) = dx/x
INT 1/(1 + y) dy = INT 1/x dx
ln(1 + y) = ln(x) + C
e^(ln(1 + y)) = e^(ln(x) + C)
1 + y = e^C*x

Let e^C = K

y = Kx - 1

Now, we have that f(-1) = 1
1 = K(-1) - 1
K = -2

y = -2x - 1

From before, we have had that x cannot equal 0. That is the only limit of the domain of this function (unless I missed something).
In other words, the domain of y is: D = (-inf,0) U (0,inf).

3. Originally Posted by Skerven
I'll post the entire problem. My math class and I couldn't figure it out. My teacher had a helluva hard time trying to explain it to us. I just want to see if anyone can understand this problem and explain it better to me. I still don't understand it
Here it goes:

"Consider the differential equation dy/dx=(1+y)/x, where x does not equal 0.

(a) On the axis provided, sketch a slope field for the given differential equation at the eight points indicated. (you can just skip this one and get to the hard question)

(b) Find the particular solution y=f(x) to the differential equation with the initial condition f(-1)=1 and state its domain."

So the last part of part b is the problem I am really asking for you to solve, although the first part of (b) as well as (a) will help.
The DE is of variables seperable type with solution y=-2x-1 for the initial
condition f(-1)=1.

Now as a function y=-2x-1 is defined everywhere on R, so in a sense its
domain is all of R. However as a solution of the DE there is a problem with
any solution as we go through x=0. The solution is not unique on any
interval that contains x=0, to show that consider:

f(x) = -2x-1 for -infty<x<0
f(x) = x-1 for 0<=x<infty

This satisfies the DE everywhere where the derivative is defined (and continuous everywhere).

The problem is that the value of a solution at -1 does not determine the
value of a solution for any positive x.

Think of it this way: Start integrating numericaly the ODE away from x=-1,
you will hit a singularity at x=0 beyond which you cannot take the solution.

So the domain of any solution of the DE satisfying an initial condition at
some negative x is (-infty,0) (as the DE does not determine the solutions
values at x>0).

RonL

4. Originally Posted by CaptainBlack
The DE is of variables seperable type with solution y=-2x-1 for the initial
condition f(-1)=1.

Now as a function y=-2x-1 is defined everywhere on R, so in a sense its
domain is all of R. However as a solution of the DE there is a problem with
any solution as we go through x=0. The solution is not unique on any
interval that contains x=0, to show that consider:

f(x) = -2x-1 for -infty<x<0
f(x) = x-1 for 0<=x<infty

This satisfies the DE everywhere where the derivative is defined (and continuous everywhere).

The problem is that the value of a solution at -1 does not determine the
value of a solution for any positive x.

Think of it this way: Start integrating numericaly the ODE away from x=-1,
you will hit a singularity at x=0 beyond which you cannot take the solution.

So the domain of any solution of the DE satisfying an initial condition at
some negative x is (-infty,0) (as the DE does not determine the solutions
values at x>0).

RonL
I've never seen something like this before. Maybe my teacher didn't explain it well enough, I wasn't paying attention, or perhaps he skipped these exceptions (though this seems like a pretty big thing to skip).

If I'm understanding you correctly, since the differential equation is undefined at x = 0 for all values of y, we cannot use an initial value starting on x < 0 to find solutions for x > 0.

Would we be able to draw the same conclusion for any solution to a differential equation that passes through some point where the differential equation is undefined? In other words, is it impossible to indicate what the particular solution will be for any domain past the point where differential equation is undefined depending upon the initial conditions?

5. Originally Posted by ecMathGeek
I've never seen something like this before. Maybe my teacher didn't explain it well enough, I wasn't paying attention, or perhaps he skipped these exceptions (though this seems like a pretty big thing to skip).

If I'm understanding you correctly, since the differential equation is undefined at x = 0 for all values of y, we cannot use an initial value starting on x < 0 to find solutions for x > 0.

Would we be able to draw the same conclusion for any solution to a differential equation that passes through some point where the differential equation is undefined? In other words, is it impossible to indicate what the particular solution will be for any domain past the point where differential equation is undefined depending upon the initial conditions?
There are bound to be some exceptions that I have forgotten, but in general
you cannot integrate an first order ODE from initial conditions on oneside of a
sigularlarity to the other. This is true for 1st order ODEs, I would have to
think about it a bit for higher order, but think this is still true.

(I don't have time at present to think through if there are any exceptions
in the case or higher order equations at present)

RonL

6. well the answer to (b) is x>0

can any one figure out why? trust me this is the right answer. my teacher tried to explain it to me.. how you have to first look at the derivative and how the derivative is bounded by y and x values..

7. Originally Posted by Skerven
well the answer to (b) is x>0

can any one figure out why? trust me this is the right answer. my teacher tried to explain it to me.. how you have to first look at the derivative and how the derivative is bounded by y and x values..
Because its wrong. A "solution" with domain x>0, cannot be a solution
to an initial value problem with an initial condition set at x=-1.

The domain (to the problem as asked here) as I said earlier is x<0.

RonL

8. Originally Posted by CaptainBlack
Because its wrong. A "solution" with domain x>0, cannot be a solution
to an initial value problem with an initial condition set at x=-1.

The domain (to the problem as asked here) as I said earlier is x<0.

RonL

How does an initial condition define a domain?
Btw... I might have missed something crucial in your post, since I don't know what "ODE" means :?

9. ## Re:

RE:

See Initial value problem - Wikipedia, the free encyclopedia

ODE means Ordinary Differential Equation...

10. Originally Posted by Skerven
How does an initial condition define a domain?
The domain MUST contain the initial value (in this case, x = -1) for a solution to the IVP to exist. To say the domain is x > 0 strictly means x cannot be negative, but any solution whose domain is x > 0 can not be a solution to the differential equation where the initial value of x in the IVP is less than 0.

In other words, either there exists a solution whose domain includes x = -1 or there is no solution to this IVP. Therefore, if the domain of y is x > 0, then y is not a solution to this IVP.

11. Originally Posted by ecMathGeek
I've never seen something like this before. Maybe my teacher didn't explain it well enough, I wasn't paying attention, or perhaps he skipped these exceptions (though this seems like a pretty big thing to skip).
I told my teacher about this problem the other day and asked him to figure out what the domain is. He went through the same steps I went through and absent-mindedly came to the same conclusion (that x != 0 was the only restriction on the domain of the solution). Then I told him that the domain was x < 0 and all of a sudden I could see a lightbold click on above his head.

It turns out my teacher did not discuss these types of exceptions in my class. He said he had gone over these in past years but because of time constraints, he skipped it this semester.

I do have one last question: Is the domain restriction in this problem strictly a result of the fact that we cannot integrate over a singularity? Or can it be argued that the absolute values in the solution (from within the ln functions) are what gave us problems when trying to find solutions to the ODE for domains crossing over x = 0? In other words, is it in part the fact that there were absolute values in the problem that forced us to limit the domain, or is that limitation strictly a result of the existance of a singularity in the ODE?

12. Originally Posted by Skerven
How does an initial condition define a domain?
Btw... I might have missed something crucial in your post, since I don't know what "ODE" means :?
See the excellent posts of qbkr21 and ecMathGeek.

To summarise the domain of the solution of an ODE (Ordinary Differential Equation,
the P is to distinguish such from a PDE a Partial Differential Equation) IVP (Initial Value
Problem) is the region over which the solution is valid. A simple requirement is that it
is at least valid at the initial value of the x or t variable, so in this case x=-1 must be
in the domain.

RonL

13. Originally Posted by ecMathGeek

I do have one last question: Is the domain restriction in this problem strictly a result of the fact that we cannot integrate over a singularity?
Yes

Or can it be argued that the absolute values in the solution (from within the ln functions) are what gave us problems when trying to find solutions to the ODE for domains crossing over x = 0? In other words, is it in part the fact that there were absolute values in the problem that forced us to limit the domain, or is that limitation strictly a result of the existance of a singularity in the ODE?
I would argue that the logs and absolute values at intermediate stages
of finding the solution are not essential to the restriction on the domain.
There may in fact be other ways of arriving at the solution where no logs
or absolute values occur.

For instance consider the IVP:

y' = y^2/x^2, y(-1)=1

this has exactly the same type of restriction on the domain of the solution
but there are no logs or absolute values in the way I would normaly solve this.

RonL

14. Can you give me a proof on how the particular solution cannot have any positive x values?

15. Originally Posted by Skerven
Can you give me a proof on how the particular solution cannot have any positive x values?
If you look at my original post you will see that the IVP does not determine
the value for a solution in this case for any x>0 since I gave a family of
functions which satisfy the IVP, but have different behaviours for x>0.

It is not that you cannot find a solution that has values for positive x, its
that such solutions are not determined by the IVP. In fact all the solutions
given can be reguarded as functions on R, its just that you can find a number
of them, but they do all agree for x<0.

RonL