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Homework Help: First-Order Equations problem

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the equation y' = f(at + by + c)
    where a, b, and c are constants. Show that the substitution x = at + by + c
    changes the equation to the separable equation x' = a + bf(x).
    Use this method to find the general solution of the equation y' = (y+t)^2

    2. Relevant equations

    n/a

    3. The attempt at a solution

    not sure where to begin :/
     
  2. jcsd
  3. Jan 14, 2010 #2

    rock.freak667

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    so x = at + by + c

    x'=?
     
  4. Jan 14, 2010 #3
    mm so plug in the first equation into where by is?
     
  5. Jan 14, 2010 #4

    Dick

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    Differentiate x with respect to t and solve for y' in terms of x'. Then put that expression for y' into your original equation. Now everything is in terms of x.
     
  6. Jan 14, 2010 #5
    thanks! i got how to do the first part, showing the substitution to show separable equation..

    How do i continue on to second part of the question?? SOrryy im kinda slow :(
     
  7. Jan 14, 2010 #6

    Dick

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    Slow is ok, but your aren't helping yet. As rock.freak667 already asked, what is x'?
     
  8. Jan 14, 2010 #7
    x' = a + by' ??
     
  9. Jan 14, 2010 #8

    Dick

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    Sure. Now solve that for y' and put it back into your original equation to eliminate y.
     
  10. Jan 14, 2010 #9
    y' = ( x' - a ) / b

    then f(x) = (x'-a)/b

    then x' = a + bf(x)
     
  11. Jan 14, 2010 #10

    Dick

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    Well, that's that then, right?
     
  12. Jan 14, 2010 #11
    still not understanding
    Use this method to find the general solution of the equation y' = (y+t)^2
     
  13. Jan 14, 2010 #12

    Dick

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    Try it. Substitute x=y+t. Do the same thing you just did to get an equation for x. What is it?
     
  14. Jan 14, 2010 #13
    x = y + t
    x' = y' +1

    y= x^2...


    x'-1= x^2
     
  15. Jan 14, 2010 #14

    Dick

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    Nice. Ok, so dx/dt=(1+x^2). That's a separable ode.
     
  16. Jan 14, 2010 #15
    integral ( -x^2 dx) = integral (1)

    - 1/3 (x^3) = t

    x = cubed root (-3t) ?
     
  17. Jan 14, 2010 #16

    Dick

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    Oh, come on, that's just silly. dx/(1+x^2)=dt. Integrate both sides. I was sort of hoping you knew separable ODE's. Heard of them?
     
  18. Jan 14, 2010 #17
    haha just started learning them :P
     
  19. Jan 14, 2010 #18

    Dick

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    Now's the time to use them.
     
  20. Jan 14, 2010 #19
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