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    Please see the attached image.
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    Firstly, you have assumed (part of) you conclusion. You are supposed to be showing that the only extremals are those satisfying y(1)=1, and y(b) = 3, not just assuming it. I'll come back and have a look at the rest of it when I'm reasonably awake.
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    (Original post by BlueSam3)
    Firstly, you have assumed (part of) you conclusion. You are supposed to be showing that the only extremals are those satisfying y(1)=1, and y(b) = 3, not just assuming it. I'll come back and have a look at the rest of it when I'm reasonably awake.
    I see what you mean, but i just have no clue on how to show this!
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    (Original post by John taylor)
    I see what you mean, but i just have no clue on how to show this!
    Transversality condition?
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    (Original post by BabyMaths)
    Transversality condition?
    I have no idea what this is. I haven't even been taught this.
    How would you go about answering this using that condition?
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    (Original post by John taylor)
    I have no idea what this is. I haven't even been taught this.
    How would you go about answering this using that condition?
    In that case there must be another way to go but here goes anyway.

    If the end point satisfies \tau(x,y)=0 then \tau_x F_{y'}+\tau_y(y'F_{y'}-F)=0 at x=b.

    In this case \tau(x,y)=y-3.

    You can read more about it on page 266 here.

    It certainly seems to work and produces the solutions that they want which turn out to be y=x and y=2-x....I think.
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    (Original post by BabyMaths)
    In that case there must be another way to go but here goes anyway.

    If the end point satisfies \tau(x,y)=0 then \tau_x F_{y'}+\tau_y(y'F_{y'}-F)=0 at x=b.

    In this case \tau(x,y)=y-3.

    You can read more about it on page 266 here.

    It certainly seems to work and produces the solutions that they want which turn out to be y=x and y=2-x....I think.
    How did you get tau? Also y(x) = x does not satisfy y(-1) = 3
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    (Original post by John taylor)
    How did you get tau? Also y(x) = x does not satisfy y(-1) = 3

    \tau(x,y)=y-3 because then \tau(b,y(b))=\tau(b,3)=0.

    With y=x you have y(1)=1 and y(3)=3. b=3.

    With y=2-x you have y(1)=1 and y(-1)=3. b=-1.
 
 
 

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