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Simple ODE, but how to approach?

  1. May 11, 2005 #1
    I'm creeping my way through DiffEq, and recently started reading Paul Dawkins' PDF, which is actually pretty helpful. He does, however, tend to assume that his readers know how to approach what he calls simple problems.

    Well, one of 'em has me stumped.

    2t*y' + 4y = 3

    I need to find the general solution. Here's what I tried:

    1) I tried isolating y. I get y=(3 - 2t*y')/4. That doesn't work because I've got y' as part of the solution. Eeek.

    2) I tried isolating y'. That didn't work either, for the same reason. I got

    y' = (3-4y)/2t

    But if I take the derivative of that, I get weirdness. Now, it's really possible that I'm just calculating the derivative incorrectly.

    I also can't figure out how to ditch the "t", but when I look at the answer, there's t, as well as C, so I know that (because of the arbitrary constant) there's a derivative involved, and that means the "t" isn't going anywhere.

    Anyone have any suggestions on how to work this? I don't need the answer. I need to know how to approach the problem.

    Thanks in advance.
     
  2. jcsd
  3. May 11, 2005 #2

    AKG

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    2ty' + 4y = 3
    2ty' = 3 - 4y
    y'/(3 - 4y) = 1/(2t)
    dy/(3 - 4y) = dt/2t
    -0.25ln(3 - 4y) = 0.5ln(t) + C
    ln(3 - 4y) = -2ln(t) + D
    3 - 4y = E/t²
    -4y = E/t² - 3
    y = F/t² + 0.75

    This is the general solution. It just uses separation of variables. F varies over R, and given an initial value, you can find the particular solution by solving for F.
     
    Last edited: May 11, 2005
  4. May 11, 2005 #3
    Thanks for the response. Now I see where I missed a step!
     
  5. May 18, 2005 #4
    Wait, I'm confused again....

    Essentially, y' is dy. I get that. But where does dt come from? This has me stumped. If I interpret the change from y' to dy as a simple change in notation, I can't just pull dt out of thin air.

    And when I perform that step, I get:

    dy/(3-4y) = 1/2t

    which yields:

    (-ln |4y-3|)/4 = (t^2)/4 + C

    Right? Then the 4s cancel out, and I'm left with

    - ln |4y-3| = t^2 + C

    But this isn't right. Where am I missing it?

    I realize this is a pretty simple separable equation, but I'm missing some crucial steps that are interfering with my understanding of the overall concept.

    Hopefully that makes sense.

    Daemon
     
  6. May 18, 2005 #5

    dextercioby

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    Nope.Rewrite it and take care with that "dt" and that "t".

    [tex] \frac{dy}{3-4y}=\frac{dt}{2t} [/tex]

    Daniel.
     
  7. May 18, 2005 #6

    arildno

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    BikDamon:
    The trick is based upon use of the integration technique "substitution", which is, essentially mirrors the chain rule of differentiation.
    Suppose you can write your differential equation as:
    [tex]f(y(t))\frac{dy}{dt}=g(t)[/tex]
    Now, integrate this between arbitrary instants "0" and "t":
    [tex]\int_{0}^{t}f(y(\tau))\frac{dy}{d\tau}d\tau=\int_{0}^{t}g(\tau)d\tau[/tex]
    where I've introduced the dummy variable [tex]\tau[/tex] for clarity (and rigour).

    Furthermore assume it exists a function F(x), so that F'(x)=f(x).
    We may then rewrite our equation as:
    [tex]\int_{0}^{t}\frac{dF}{dx}\mid_{x=y(\tau)}\frac{dy}{d\tau}d\tau=\int_{0}^{t}g(\tau)d\tau[/tex]
    That is, we may write this as:
    [tex]\int_{0}^{t}(\frac{d}{d\tau}F(y(\tau)))d\tau=\int_{0}^{t}g(\tau)d\tau[/tex]
    by the chain rule of differentiation.
    But the left hand side equals now by FOTC:

    [tex]\int_{0}^{t}(\frac{d}{d\tau}F(y(\tau)))d\tau=F(y(t))-F(y(0))[/tex]
    which equals, with [tex]y(0)=y_{0}, y(t)=y:
    [tex]F(y(t))-F(y(0))=\int_{y_{0}}^{y}\frac{dF}{dy}dy=\int_{y_{0}}^{y}f(y)dy[/tex]
    Thus, we have gained the equality:
    [tex]\int_{y_{0}}^{y}f(y)dy=\int_{0}^{t}g(\tau)d\tau[/tex]
     
  8. May 18, 2005 #7
    I get the theory...

    I understand the theory in the last response, but I got lost on the actual application.

    Essentially, once I replace y' with dy, I arbitrarily assign dt to the right side of the equation, yes?
     
  9. May 18, 2005 #8

    dextercioby

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    [tex] y'\equiv \frac{dy}{dt} [/tex].You could (even though it's mathematically doubtful) move around those differentials (divide by them,multiply by them).

    Daniel.
     
  10. May 18, 2005 #9

    arildno

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    You do NOT replace y' with dy; you replace y'dt with dy !!!!!!
     
  11. May 18, 2005 #10

    dextercioby

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    And one more thing:nothing you do here is arbitrary.Only integration constants have that privilage.

    Daniel.
     
    Last edited: May 18, 2005
  12. May 18, 2005 #11

    arildno

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    And, read again my prior response.
    Make sure you undestand it properly.
     
  13. May 18, 2005 #12
    I think part of my challenge is that the originally stated problem does not include "dy" or "dt" and therefore working towards the solution has become a bit convoluted for me.

    The original problem only says: 2ty' + 4y = 3

    Do you guys even see where I'm getting lost? If not, I'll move on to the next problem!
     
  14. May 18, 2005 #13
    so the problem doesn't specify that y is a function of t? well that sure can be misleading.
     
  15. May 18, 2005 #14

    dextercioby

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    You're getting lost at this step [tex] y'\equiv \frac{dy}{dx} [/tex],which is really sad.

    Daniel.
     
  16. May 18, 2005 #15

    Pyrrhus

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    I believe that what Daniel refers as "that's sad" is the fact that you should know Leibniz notarion by now, way before taking a course on differential equations.
     
  17. May 18, 2005 #16
    Gotcha. Actually, I *have* worked with the notation before. What I didn't realize in the problem was that the notation was implied. The problem came from a summary of term definitions in a larger document, and I'm working my way through it.

    I'm not trying to belabour the issue; I just wanted to make sure that I understood exactly what was going on.

    Thanks.
     
  18. May 18, 2005 #17

    arildno

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    It's perfectly common in the beginning to be confused by different notations.
    Hopefully, things have cleared out now.
     
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