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Thery question

  1. Jun 28, 2009 #1
    Hello, :smile:


    I might sound odd, but I need to understand how you come up with a differential equation then end it up with some solution.

    How sure are you that your solution is correct ? :biggrin:


    THANK YOU :smile:
     
  2. jcsd
  3. Jun 28, 2009 #2

    nicksauce

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    It is easy to show that a solution to a differential equation is correct. Simply substitute your solution into the differential equation and show that it indeed solves the equation.
     
  4. Jun 28, 2009 #3
    But how do you make up an equation ?
     
  5. Jun 28, 2009 #4
    Start with a function, and take it's derivative e.g.

    y = sqrt(x)

    y' = 1/{2 sqrt(x)}

    Therefore we have the differential equation:

    y' = 1/{2 y}

    I think it is good practice for students to do these types of exercises.
     
  6. Jun 29, 2009 #5
    If you are referring to ways to formulate a differential equation from a physical application, calculus books usually give some of the easy basic examples.

    If you were given two variables: x, y; and you were told that y is proportional to x, how would you come up with an equality to relate the two variables? You would probably come up with something similar to (for constant of proportion k):

    [tex]y=kx[/tex]

    Now, suppose you are given a function y(x) and told that its rate of change is proportional to its position for every value of x. What type of equality would you find for this? Something like:

    [tex]\dfrac{dy}{dx}=ky[/tex]

    It's that easy in this example, all we did was find an equality that relates the function and its derivative based on our given information. Solving this ODE gives us the formula for exponential growth (or decay, for negative values of k) at time x. Perhaps a variable t would have been a better choice. Nevertheless:

    [tex]\dfrac{dy}{dx}=ky[/tex]

    [tex]\displaystyle\int\frac{dy}{y}=\displaystyle\int kdx[/tex]

    [tex]\ln(y)=kx+c[/tex]

    [tex]y(x)=e^ce^{kx}=y_0e^{kx}[/tex]
    because if we set x=0 then we have y(0)=e^c and can reassign that constant to e^c=y_0.

    Perhaps later if I feel more creative and awake I can come up with a better example of formulating a differential equation. Many, many, others are certainly not this easy.
     
  7. Jun 29, 2009 #6

    HallsofIvy

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    Many differential equations from physics problems come from "force equals mass times acceleration". Acceleration is the second derivative of position so that gives a second order differential equation:
    [tex]m\frac{d^2x}{dt^2}= F(x, t)[/tex]
    where F(x,t) is the function giving the force on the object at each position x and time t.

    In many chemistry problems you are given the rate at which chemicals combine which immediately leads to a differential equation.

    But in general, HOW you derive a differential equation from a problem (or whether a differential equation relates to the problem at all) depends strongly upon the problem.
     
  8. Jun 29, 2009 #7

    quasar987

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    There are several ways. For instance

    (1) From experimental data. The derivative of a quantity Q represents the speed at which Q varies. One might measure experimentally the relation between the derivatives of various quantitie and the mathematical way to express this relation is a differential equation.

    (2) From another differential equation. The wave equation is a differential equation that is obtained from Newton's 2nd law a motion (which is itself a differential equation) by making some suppositions and approximations.

    But note that as with most (every?) equations of physics, the solutions are only approximations to reality, so while they are correct mathematically, they only approximate the physics (but often astonishingly well!).
     
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