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ODE question

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    show the initial value problem x(dy/dx)=4y, y(0)=1 has no solution. does this contradict the existence theorem. please explain

    3. The attempt at a solution

    it is easy to find out a general solution is y=C*e^(4x), C is a constant. and for any x the right part of the equation will always be
    ~~~~I'm sorry, I made a mistake here, the general solution should be y=C*x^4

    bigger than zero, so there is no solution for y(0)=-1, the question is how to explain it?

    the existence theorem I learned from class is the following:
    For F(t,y,y')=0, I.C. y(x0)=y0 ...(EQ1)
    Let R be the region a<x<b, c<y<d. Such that (x0,y0) belongs to R. If f(x,y) is continuous and bounded on R, then EQ1 has a solution. The validity of the solution is in R.

    Last edited: Nov 16, 2008
  2. jcsd
  3. Nov 16, 2008 #2


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    Homework Helper

    [tex]\int \frac{1}{y} dy=\int \frac{4}{x}dx[/tex]

    Your general solution is incorrect. The differential eq'n can be seperated as shown above. Integrate again and simplify. Then put in the initial values and see if a solution exists.
  4. Nov 16, 2008 #3
    I agree the poster's solution is faulty...The final result should be y=B(x^4),but that also shows there is no solution fo y(0)=1!Because with 1=B*0 you cant find the solution for constant B....THERE IS NO SOLUTION AS SHOWN ABOVE,mathematically the language is understood,this part am sure of!...Now about the EXISTENCE THEOREM[this part am not so sure,its logical and could be anything,I just read it anyways]...there is no clear statement associated with the question,I dont see an EXISTENTIAL QUANTIFIER to count for anything...ALTHOUGH,abstractly I would consider that the statement "show that there is no solution for y(0)=1" is equivalent to an existential quantifier of statement "for a value x=0 there is no value y=1" which DOES NOT really contradict the existence theorem...I still have doubts and this is not something to completely count on though I hope it has been helpful!
  5. Nov 16, 2008 #4
    you are right, i made a mistake. I changed the answer.

    after put the initial conditions, there is still no solutions.

  6. Nov 16, 2008 #5
    thanks for your reply!

    I posted the existence theorem I learned from class.

    In my opinion, the function we get y=Ax^4 is continuous, however, by using the existence theorem we could not find a solution. that is the contradict point.

    maybe the existence theorem is just sufficient. maybe that is why?
  7. Nov 17, 2008 #6


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    Staff Emeritus
    Science Advisor

    You quoted the existence theorem before as saying "If f(x,y) is continuous and bounded on R, then dy/dx= f(x,y) has a solution."

    What is f(x,y) for this problem?
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