Results 1 to 15 of 15

Math Help - Crazy Domain Problem from an AP Exam

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    34

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Skerven View Post
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Skerven View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by CaptainBlack View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ecMathGeek View Post
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2007
    Posts
    34
    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..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Skerven View Post
    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2007
    Posts
    34
    Quote Originally Posted by CaptainBlack View Post
    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 :?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    RE:

    See Initial value problem - Wikipedia, the free encyclopedia

    ODE means Ordinary Differential Equation...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Skerven View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by ecMathGeek View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Skerven View Post
    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
    Last edited by CaptainBlack; May 6th 2007 at 01:11 AM.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ecMathGeek View Post

    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
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    May 2007
    Posts
    34
    Can you give me a proof on how the particular solution cannot have any positive x values?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Skerven View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Please help this problem is driving me crazy
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: January 30th 2010, 04:45 PM
  2. Crazy problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 12th 2009, 02:26 PM
  3. Crazy parametric problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 29th 2009, 03:35 PM
  4. integer problem (please help me! Im going crazy!)
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 28th 2008, 10:55 AM
  5. Crazy percent problem
    Posted in the Math Topics Forum
    Replies: 9
    Last Post: January 19th 2007, 11:53 AM

Search Tags


/mathhelpforum @mathhelpforum