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Extremely Basic Differential Equations

  1. Apr 5, 2010 #1

    Char. Limit

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    My AP Calculus class touched on differential equations recently, only for a moment. I immediately noticed that most differential equations that involved only multiplication were easy. However, I was soon stumped by some addition differential equations. How do you solve these undoubtedly bonehead simple equations?

    (Note: As the work is being done independently, it isn't homework, and as such I'm not sure if I need to show what I've done so far. Besides, I don't know much at all about differential equations. I'd have nothing to show.)

    [tex]\frac{dy}{dx}=x+y[/tex]

    [tex]\frac{dy}{dx}=x-y[/tex]

    [tex]\frac{dy}{dx}=y-x[/tex]

    And generalizations...

    [tex]\frac{dy}{dx}=ax+by[/tex]

    [tex]\frac{dy}{dx}=ax-by[/tex]

    [tex]\frac{dy}{dx}=ay-bx[/tex]

    for constant real a and b.
     
  2. jcsd
  3. Apr 5, 2010 #2

    rock.freak667

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    [tex]\frac{dy}{dx}+P(x)y=Q(x)[/tex]


    a DE in this form can be solved by multiplying by an integrating factor u

    [tex]u=e^{\int P(x) dx}[/tex]


    the left side can then be written as d/dx(yu)


    EDIT: Yep, Dick is right...it'd be overkill with the method above.
     
    Last edited: Apr 5, 2010
  4. Apr 5, 2010 #3

    Dick

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    These look like problems where you are better off doing substitution. Try u=x+y for the first one. Solve for dy/dx in terms of du/dx.
     
  5. Apr 5, 2010 #4

    Char. Limit

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    So, for the first one, I did this...

    [tex]u=x+y[/tex]

    [tex]\frac{du}{dx}=1+\frac{dy}{dx}[/tex]

    [tex]\frac{dy}{dx}=\frac{du}{dx}-1[/tex]

    [tex]\frac{du}{dx}-1=u[/tex]

    [tex]\frac{du}{dx}=u+1[/tex]

    [tex]\frac{du}{u+1}=dx[/tex]

    [tex]ln(u+1)=x+C[/tex]

    [tex]u+1=C e^x[/tex]

    [tex]x+y+1=C e^x[/tex]

    [tex]y= C e^x - x - 1[/tex]

    But is that right? It seems like a ridiculously complicated answer for such a simple equation...

    I'm working on the second one now.
     
    Last edited: Apr 5, 2010
  6. Apr 6, 2010 #5

    Mark44

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    Differential equations can sometimes be difficult to solve, but they are much easier to check. Take the derivative of your solution to the first problem to see if you get dy/dx = x + y.
     
  7. Apr 6, 2010 #6
    Surely the first two equations are wrong?

    If you take [tex]ln(u+1)=x+C[/tex]

    then using exponents would transform these to [tex]e^{ln(u+1)}=e^{x+C}[/tex]

    which then cancels to

    [tex]u+1=e^{x} \cdot e^{C}[/tex]

    Usually in differential equations you take [tex]e^{C}[/tex] to be a constant [tex]A[/tex]

    So we end with the final equation

    [tex]y=Ae^{x}-(x+1)[/tex]
     
  8. Apr 6, 2010 #7

    Char. Limit

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    Well, I consider the constant to be any arbitrary real number, given no initial conditions.

    So, a real number raised to the power of an arbitrary real number is another real number...

    So C and e^C are both arbitrary real numbers, and can be expressed by a constant. The letter still stands for something, and if you solved for C_1 or C_2 you'd get the same equation, so why worry?
     
  9. Apr 6, 2010 #8

    rock.freak667

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    I think you solved it correctly, you just didn't really put any separation with the lines so it would be difficult to read.
     
  10. Apr 6, 2010 #9
    My point was that you may not always get away with just 'skipping' steps/information.

    Even if you did do the steps mentally, you should really change your intercept value to [tex]d[/tex] or just something other than [tex]C[/tex]
     
  11. Apr 6, 2010 #10

    Char. Limit

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    I suppose you're right...

    I'll remember that from now on.
     
  12. Apr 8, 2010 #11

    Char. Limit

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    DEQs revisited...

    On my first AP Calculus test, one of the problems involved the differential equation [itex]\frac{dy}{dx}=2x-y[/itex] with initial condition [itex]y(0)=1[/itex]. Needless to say, using the substitution method you guys taught me, I got the answer ([itex]y=3e^{-x}+2x-2[/itex]). So, the next one really threw me for a loop...

    The second practice AP test had the first-order nonlinear ODE...

    [tex]\frac{dy}{dx}=5x^2-\frac{6}{y-2}[/tex]

    I don't even know where to start on this. It's not even an exact equation... and I tried to turn it into one using a mathematical dictionary, but it didn't quite work, or maybe I'm not understanding how to turn an inexact FONODE into an exact FONODE. Is this unsolvable?
     
  13. Apr 8, 2010 #12

    Mark44

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    There might be an integrating factor you can find that will turn this into an exact DE. If you're studying differential equations, this technique should be in your book.
     
  14. Apr 8, 2010 #13

    Char. Limit

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    Well, that's the problem. I'm not in a DE class. Everything I'm learning about DEs I'm learning from a Concise Dictionary of Mathematics, and I thought it would be enough, but I can't figure out how to get the integrating factor.
     
  15. Apr 8, 2010 #14

    Mark44

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  16. Apr 8, 2010 #15

    Char. Limit

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    OK, so let's try to apply that to the above equation...

    Start with...

    [tex]\frac{dy}{dx}=5x^2-\frac{6}{y-2}[/tex]

    OK, so let's add 6/(y-2) to both sides, I think...

    [tex]\frac{dy}{dx} + \frac{6}{y-2} = 5x^2[/tex]

    So I have a Q(x) on the right, but I need a P(x)y on the left. I guess multiply top and bottom of 6/(y-2) by y.

    [tex]\frac{dy}{dx} + \frac{6}{y^2-2y} y = 5x^2[/tex]

    Now, that looks more like a P(y)y, but I'm going to try not to let that stop me...

    Or, maybe I will...

    How do I change the [itex]\frac{6}{y^2-2y}[/itex] into something I can integrate wrt x?
     
  17. Apr 8, 2010 #16

    Mark44

    Staff: Mentor

    On closer look, I don't see my suggestion of using an integrating factor as being helpful with this DE. I don't have any ideas on this one.

    I've posted this problem in the Homework Helpers section, so maybe someone will have a good idea.
     
    Last edited: Apr 8, 2010
  18. Apr 8, 2010 #17

    vela

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    Try the substitution u=1/(y-2). It looks promising.

    Edit: Well, maybe not...
     
    Last edited: Apr 8, 2010
  19. Apr 8, 2010 #18

    Are you sure you were supposed to solve that DE to find the answer to whatever the question asked? That seems like a pretty complicated DE to solve for a class not about DE...
     
  20. Apr 8, 2010 #19

    Char. Limit

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    No, it could be... I tried it anyway, and I did this:

    Start with...

    [tex]u=\frac{1}{y-2}[/tex]

    Differentiate both sides...

    [tex]\frac{du}{dx}=-\frac{1}{(y-2)^2}\frac{dy}{dx}[/tex]

    Substitute u for 1/(y-2)...

    [tex]\frac{du}{dx}=-u^2\frac{dy}{dx}[/tex]

    And solve for dy/dx...

    [tex]-\frac{1}{u^2}\frac{du}{dx}=\frac{dy}{dx}[/tex]

    And we can now put in the u values into our original equation...

    [tex]\frac{dy}{dx}=5x^2-\frac{6}{y-2}[/tex]

    [tex]-\frac{1}{u^2}\frac{du}{dx}=5x^2-6u[/tex]

    Which can be rewritten as...

    [tex]\frac{du}{dx}=6u^3-5x^2u^2[/tex]

    Don't know what to do with that though.
     
  21. Apr 8, 2010 #20

    Char. Limit

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    The main problem doesn't ask to solve it, only to approximate it. However, the challenge problem of this week and next week (he issues challenges sometimes) is to solve it.
     
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