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(probably stupid) Differential equation question

  1. Feb 21, 2009 #1
    Hey all, i just lost a TON of points on a test for solving a

    problem in a way that is apparently invalid.

    The problem was verify that y(x) = x+1 is a solution for dy/dx =

    y*y-x*x-2x; y(0) = 1. i plugged y = x+1 into the right side of

    the second equation, got dy/dx = 1, integrated to get y=x+c, used

    y(0) = 1 to get c= 1, therefore y = x + 1

    My professor's annoyed 2-second explanation about why my method

    is invalid was that I assumed that it worked to prove that it

    worked. I sorta buy it, but I'm not completely convinced, could

    someone give me a counter example to prove that my method is not

    legit? [to clarify, my method is to plug in y(x) into the DE,

    then integrate, then use the given initial conditions to solve for

    c to get a new y(x) and make sure that my new y(x) is the same as

    the old one].

    The counter example I am requesting would take a form that is

    similar to the problem above, except that y(x) would not be a

    legit solution to dy/dx, BUT my method would falsely show that

    y(x) does work. Obviously, if no such counter example exists,

    that my method proves that the DE works and I should not have lost

    any points

    Thanks in advance!
     
  2. jcsd
  3. Feb 22, 2009 #2

    Mark44

    Staff: Mentor

    What you did was sort of a cross between verifying that a given function was a solution and attempting to find the solution.

    When you substituted y = x + 1 into the right side to get 1, why didn't do the same substitution on the left side? After all, if y = x + 1, dy/dx = 1.

    When you got dy/dx = 1, that's not the same differential equation as the one you started with. dy/dx happens to be equal to 1 when y = x + 1. By treating dy/dx as a constant, you are eliminating all of the other potential soltutions of the DE dy/dx = y^2 - x^ - 2x. Off the top of my head I don't know what the other solutions to this DE might be, but such an equation (without the initial condition) generally has an infinite number of them.

    There's a big difference between being asked to verify that a function is a solution of an initial value problem (a DE + a set of initial conditions), and finding the solutions to a DE. At this point, you probably don't have the tools to solve nonlinear DEs like this one, so make your life a little easier and do what is asked for. After you've done that, you can explore alternate techniques.
     
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